^^/^(/. 


Sheldon  &  Co77ipany's  Text-Hooks, 

_  _ 

The  Scieiwe  of    Government  in    Connection  tvith 
American  Institutions,    By  Joseph  Alden,  D.D.,  LL.D., 

Pres.  of  State  Normal  School,  Albany.   1  vol  12mo. 
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Hereafter  no  American  can  be  said  to  be  educated^  who  does  not  thoroughly 

ui    ------      -  ^  —  jjaid, 

th  ;hose 

de     S    "    ""    *"~'* 

-I  PKIVATELIBEAEY 

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h^  ^x«o«jixv,vi.  ttiiv.!.  axxuiigcu.  0.0  \j\j  jiauiiitait;  iiiu  exinessiori  of 
ideas,  and  assist  in  literary  composition.  By  PetePv  Mark 
ROGET.  Revised  ^1  edited,  with  a^Ljst  of  Foreigii  \Vor^s 
defined  in  EnglTsh,  and  other  additions,  by  Barn  as  Sears,  D.D., 
late  President  of  Brown  University.  A  new  American,  from 
the  last  London  edition,  with  important  Additions,  Corrections, 
and  Improvements.     12mo,  cloth. 

Faircliilds'  Moral  rUilosophy ;  or,  The  Science  of 


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doneTTEFHoctrine  that  virtue,  in  its  elementary  form,  consists  m  benevo- 
lence, and  that  all  forms  of  virtuous  action  are  modifications  of  this  principle. 

After  presenting  this  view  of  obligation,  the  author  takes  up  the  questions  of 
Practical  Ethics,  Government  and  Personal  Rights  f.nd  Duties,  and  treats 
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University  of  California  •  Berkeley 

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Early  American  Mathematics  Books 


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OLNEY'S    MATHEMATICAL     SERIES. 


THE 


COMPLETE  ALGEBEA 


EMBRACING 

SIMPLE    AND    QUADRATIC    EQUATIONS,    PROPORTION    AND 

THE    PROGRESSIONS,     WITH    AN     ELEMENTARY    AND 

PRACTICAL  VIEW  OF   LOGARITHMS;    A   BRIEF 

TREATMENT  OF  NUMERICAL    HIGHER 

EQUATIONS. 


AND    A     CHAPTER     ON     THE 

BUSINESS   RULES  OF  ARITHMETIC 

TREATED     ALGEBRAICALLY 


Designed  to  be  sufficiently  elementary  for  beginners,  and  sufficiently 
thorough  and    comprehensive  to  meet  the  wants  of  our   better 

high     schools,     ACADEMIES,     AND     ORDINARY     COLLEGES. 


BY 

EDWARD    OLNEY, 

Pro/tssor  o/  Mathematics  in  the  University  0/  Michigan. 


NEW  YORK : 

SHELDON     &     COMPANY, 


8    MURRAY    STREET. 


OLNEY'S  MATHEMATICAL  SERIES 

EMBRACES  THE  FOLLOWIifG  BOOKS. 


"Primary  Arithmetic, 

Ji7e9ne?its  of  A.rithm^etic, 

Practical    Arit?imetic, 

Teacher's  Mand-^ook  of 

Arithmetical    ^Exercises, 

Science  of  Arithmetic. 

Send  for  fiill  Circnlar  of  Olney's  Arithmetics. 

fntroduction  to  Algebra, 
Complete  Algebra, 

JECey  to   Complete  Algebra, 
University  Algebra, 

£'ey  to   U'72ire7^sity  Algebra, 

2'est  Examples  in  Algebra, 
Elements  of  Geometry, 
(Separate.) 
JSleme7its  of  Trigo7iometry, 
(Separate.) 

introduction  to   Geometry,    ^art  I, 

(Bound  separate.) 
^lemeiits  of  Geometry  and   Trigonometry, 

(Bound  in  one  Vol.) 

Geometry  a^id  Trigonometry , 
(University  Edition.) 
General   Geometry  a7id  Calculus, 


Copyright ^^  1870,  1875,  1878,  by  Sheldon  &f>  Co. 


Blectrotyped  by  Smith  &  McDouqal,  82  Beekman  St.,  N.  Y. 


PREFACE. 


Theee  main  purposes  have  controlled  the  author  in  the  preparation 
of  this  work  : — First,  to  provide,  in  a  single  volume,  of  convenient 
size,  an  elementaiy  treatise  upon  Algebra,  sufficiently  simple  for  the 
use  of  beginners  in  the  science— sufficiently  full  on  the  subjects  em- 
braced to  render  their  rediscussion  in  a  subsequent  volume  unnecessary, 
and  to  afford  a  range  of  topics  comprehensive  enough  to  meet  the 
wants  of  our  Common  and  PubUc  High  tichools,  and  for  a  good  pre- 
paration to  enter  any  of  our  Colleges  or  Universities.  Second,  to  train 
the  pupil  to  methods  of  reasoning,  rather  than  in  mere  methods  of 
operating.  Third,  to  present  the  whole  in  such  a  form  as  to  make  the 
book  a  convenient  one  to  use  in  the  class  room. 

As  the  mathematical  studies  are  at  present  pursued  in  our  schools, 
pupils  are  expected  to  have  a  fair  knowledge  of  Mental,  and  some 
knowledge  of  what  is  variously  styled  Practical  or  "Written  Arithmetic, 
before  entering  upon  the  study  of  Algebra.  Without  expressing  any 
opinion  upon  the  propriety  of  this  course,  the  author  has  accepted  it 
as  a  fact  likely  to  exist  for  some  time  to  come,  and  has  attempted  to 
adapt  his  Algebra  to  it.  This  book  is,  therefore,  not  a  mere  child's 
book,  but  is  designed  as  a  First  Book  in  Algebra  for  pupils  who  have 
the  knowledge  of  Arithmetic  usually  considered  requisite  to  a  com- 
mencement of  this  study. 

The  volume  contains  a  more  than  ordinarily  thorough  and  full  treat- 
ment of  the  processes  of  Literal  Arithmetic,  viz.,  the  Fundamental 
Kules,  Fractions,  Involution,  Evolution,  and  the  Calculus  of  Eadicals. 
It  is  in  these  processes  that  pupils  usually  find  the  most  difficulty; 
and  in  them  comparatively  few  become  proficient,  especially  in  the 
Calculus  of  Eadicals.  Yet  no  one  can  become  an  algebraist,  and,  if 
not  an  algebraist,  not  a  mathematician,  without  being  perfectly  mas- 
ter of  these  processes.  Hence  they  are  treated  so  much  at  length  and 
with  so  much  care  and  thoroughness.  It  is  idle  to  deceive  the  pupil 
with  the  notion  that  he  is  mastering  the  subject  of  algebra  by  obtain- 
ing a  superficial  acquaintance  with  the  nature  and  uses  of  some  of  the 
more  simple  forms  of  the  equation.  The  pupil  who  cannot  attain  a 
good  degree  of  familiarity  \^dth  the  Hteral  arithmetic,  cannot  appre- 
ciate mathematical  reasoning,  or  profit  much  by  this  study. 


VI   ■  PREFACE. 

The  treatment  of  Simple  and  Quadratic  Equations  will  be  found  as 
full  as  is  requisite  for  the  foundation  of  a  good  mathematical  educa- 
tion. It  is  not  the  author's  purpose  to  rediscuss  the  themes  here  con- 
sidered in  his  second  volume.  The  matter  presented  on  these  subjects 
does  not  differ  much  from  what  is  usually  contained  in  our  so-called 
higher  algebras  ;  and,  in  fact,  is  more  thoroughly  treated  than  in  most 
of  them.  Katio,  Proportion,  and  the  Progressions  have  received  a 
proper  share  of  attention.  The  transformation  of  a  Proportion  will 
be  found  to  be  discussed  in  a  somewhat  different  manner  from  the 
usual  one,  and,  it  is  hoped,  in  such  a  way  that  the  pupil  who  enters 
into  the  spirit  of  the  treatment,  and  becomes  familiar  with  the  meth- 
ods, will  feel  that  this  elegant  and  important  mathematical  instrument 
is  his  own. 

It  is  thought  that  the  chapter  on  the  Business  Rules  of  Arithmetic 
will  be  particularly  acceptable  to  a  large  class  of  students.  Some 
pr£)blems  in  Interest,  Common  Discount,  and  Alligation,  which  are 
ordinarily  considered  intricate,  will  be  found  quite  within  the  reach  of 
ordinary  minds.  Other  applications  of  the  equation  to  the  solution  of 
practical  questions,  are  not  less  extended  than  in  our  common  treatises. 
The  chapter  on  Logarithms  will  give  the  student  a  clear  conception  of 
the  nature  of  this  kind  of  number,  and  enable  him  to  understand  and 
use  the  common  tables. 

No  attempt  is  made  at  the  Discussion  of  Equations,  or  the  Devel- 
opment of  Series.  To  discuss  an  equation  well,  is  a  mathematical 
accomplishment.  It  is  not  a  thing  which  the  tyro  can  master.  He 
may  follow  another  through  such  a  discussion  ;  but  he  will  do  little 
more.  The  Development  of  Functions  is  reserved  for  the  second 
volume,  where  the  Binomial  Formula,  and  the  Logarithmic  Series  will 
be  seen  to  be  but  deductions  from  a  more  general,  and  more  easily 
produced  formula.  The  Binomial  Formula  is  given  (though  not  pro- 
duced) in  this  treatise,  and  much  care  is  bestowed  in  showing  how  to 
use  it  in  developing  the  common  forms  to  which  it  is  appUcable.  It  is 
thought  that  these  topics,  with  the  discussion  of  them  given  in  the 
text,  will  meet  the  author's  first  purpose. 

The  attempt  to  accomplish  the  second  end,  viz.,  to  train  the  pupil 
to  methods  of  reasoning,  rather  than  in  mere  methods  of  operating, 
has  given  character  to  the  presentation  of  every  topic.  Propositions 
are  clearly  stated  at  the  outset,  and  demonstrations  are  given  in  form, 
and  with  the  rigor  of  a  geometrical  argument.  That  there  is  some 
defect  in  our  methods  of  instruction,  in  this  regard,  must  be  pftinfully 
evident  to  every  one  who  has  been  called  to  examine  large  numbers  of 
our  youth  in  this  study.     The  author  has  examined  for  admission  to 


PREFACE.  Vll 

college,  from  25  to  150  diflferent  students  from  all  parts  of  our  country, 
each  year,  for  the  last  16  years,  and  he  has  almost  invariably  found 
Httle  or  no  knowledge  of  the  processes  as  arguments,  even  when  a 
good  degree  of  skill  in  the  use  of  the  processes  had  been  attained. 
Perhaps  a  majority  of  those  examined  could  multiply  the  square  root 
of  2  by  the  cube  root  of  3,  but  scarce  one  in  50  could  develop  the 
process  in  a  logical  form,  or,  in  most  cases,  give  any  rational  account 
of  it.  Now,  it  need  not  be  said  that,  in  a  course  of  education,  this  is 
a  fundamental  defect ;  it  is  failure  just  where  success  is  vital.  The 
processes  of  a  mathematical  science  are  of  comj)aratively  little  worth 
to  a  great  majority  of  those  who  study  them  ;  the  development  of  tho 
reasoning  powers  to  which  such  studies  are  addressed,  is  of  the  high- 
est importance  to  all.  By  teachers  who  cannot  appreciate  these  truths, 
this  book  will  very  probably  be  misunderstood  ;  but  to  such  as  do  feel 
the  force  of  them,  the  author  appeals  with  the  fullest  confidence,  not 
indeed  that  his  book  will  meet  the  exigency,  but  that  it  will  be  wel- 
comed as  an  effort  in  the  right  direction,  and  as  a  help  in  remedying 
this  radical  defect. 

To  such  as  use  the  book  the  author  commends  the  following 

SUGGESTIONS  TO  TEACHEKS. 

Great  pains  have  been  taken  to  adapt  the  whole  to  the  use  o£  the 
class  room.  Definitions  and  all  Propositions  are  made  to  stand  out 
clearly,  and  have  been  written  -with  great  care.  These  should  be  mem- 
orized by  the  student,  as  well  as  thoroughly  comprehended.  It  is 
also  recommended  that  the  Bules  be  committed  to  memory,  verbatim  ; 
not,  indeed,  to  be  recited  as  a  mere  parrot-like  performance,  but  as  a 
means  of  acquiring  the  language  of  the  science,  and  attaining  faciHty 
in  clothing  its  thoughts  in  a  becoming  garb.  But  let  the  Demonstra- 
tion of  every  Proposition  and  Fade  he  inseparably  connected  icith  the  state- 
ment of  the  truth  or  process.  The  author  has  endeavored  to  make  these 
demonstrations  to  their  propositions  just  what  the  demonstrations  in 
Geometry  are  to  theirs  ;  and  the  method  of  use  should  be  the  same. 

The  Model  Solutions  are  samples  of  what  the  pupil  should  give  at 
the  blackboard  in  the  class  room,  repeating  the  reasoning  in  the  * '  expla- 
nation "  of  every  example,  till  it  is  perfectly  familiar.  These  solutions 
are  not  mere  statements  of  how  the  thing  is  done,  but  are  designed  to 
be  a  logical  presentation  of  the  process  as  an  argument.  It  is  just  at 
'this  point  that  the  great,  and  well-nigh  universal  failure  referred  to 
above  occurs.  By  a  close  adherence  to  the  plan  here  presented,  it  is 
hoped  something  may  be  done  to  remedy  the  e\'il. 

The  Synopses  should  be  made  the  basis  of  reviews.     No  student 


VUl  PBEFACE. 

should  be  allowed  to  think  he  has  mastered  a  subject,  till  he  can  go  to 
the  blackboard,  put  down  the  Synopsis,  and  discuss  the  whole 
theme  therefrom,  stating  every  principle,  in  good  language,  and  dem- 
onstrating, in  good  style,  every  proposition. 

The  Test  Questions  are  not  designed  to  be  full  or  exhaustive  ;  they 
are  usually  only  a  few  detached  questions,  such  as,  if  well  answered, 
would  satisfy  an  examiner  that  a  student  understood  the  subject  under 
consideration. 

The  Examples  are  more  than  ordinarily  numerous,  and  have  been 
selected  from  the  common  sources.  In  the  Key  there  will  be  found  a 
very  large  number  of  additional  examples  for  the  teacher's  use  in  class 
room  drill.  It  will  often  be  found  serviceable  to  assign  these  in  the 
class  room  in  preference  to  those  in  the  text,  for  reasons  which  will  be 
apparent  to  every  teacher. 

As  to  originahty,  though  nothing  may  have  been  developed  which  is 
absolutely  new,  there  are  many  features  to  which  attention  might  be 
called.  Among  these  are  the  exposition  of  the  doctrine  of  the  signs  -f- 
and  — ,  the  theory  of  signs  in  multipUcation,  the  treatment  of  subtrac- 
tion, some  things  in  factoring,  and  the  whole  spirit  of  the  treatment  of 
the  Calculus  of  Radicals,  the  exposition  of  the  nature  of  a  Literal  Frac- 
tion, the  treatment  of  Proportion,  etc.  The  very  idea  of  the  character 
and  province  of  the  science,  as  given  in  the  definition  of  Algebra,  may 
not  be  without  interest.  But  the  Author  is  in  no  fear  of  being  called 
a  plagiarist.  That  he  has  not  borrowed  more  freely,  may  be  his  chief 
misfortune. 

Valuable  suggestions  have  been  received  from  many  practical  teach- 
ers ;  and  the  Author  feels  under  special  obligation  to  Mr.  Thos.  Hunter, 
President  of  the  Female  Normal  and  High  School,  New  York  City,  who 
has  read  the  entire  work  in  manuscript,  and  whose  words  of  criticism, 
not  less  than  his  generous  appreciation  and  approbation,  have  done 
much  to  foster  the  hope  that  the  work  may  not  be  without  merit. 

With  these  remarks,  the  Author  commits  his  effort  to  the  judgment 
of  his  fellow  teachers,  not  indeed  without  the  keenest  sense  of  its  im- 
perfections, but  with  the  humble  hope  that  it  may  be  found  suggestive, 
and  to  some  extent  serviceable  in  promoting  a  much  needed  reform  in 
our  methods  of  teaching  this  fundamental  branch  of  mathematical 
science. 

EDWARD  OLNEY. 

TJNrVEBSITT  OF  MICHIGAN, 

Ann  Arbor,  January,  1870. 
N.  B. — For  scope  and  purpose  of  tlio  Appendix,  soe  foot-note,  p.  3i)l. 


CONTENTS. 


INTROD  UCTION, 

SECTION  I. 

A  BRIEF  SURVEY  OF  THE  OBJECTS  OF  PURE  MATHEMAT- 
ICS AND  OF  THE  SEVERAL  BRANCHES. 

PAGE 

PxJEE  Mathematics. — Definition  {1);  branches  enumerated 
{2>3) 1 

Quantity. — Definition  (4) 1 

Number. — Definition  (5);  Discontinuous  and  continuous  (6, 
7,8) 2,3 

Definition  OF  THE  Several  Branches. — Arithmetic  \0)\  Al- 
gebra {10) ;  Calculus  {11) ;  Geometry  {12) 3,  4 

Synopsis 4 

SECTION  11. 

L0GIC0-MATHE3kL\TICAL  TERMS. 

Proposition. — Definition  (13) 5 

Varieties  of  Propositions. — Enumerated (J4);  Axiom  {15)', 
Theorem  {16);  Lemma  {18);  Corollarj'  {10);  Postulate 
{20);  Problem  {21) 5-7 

Definition  OF —Demonstration  {17);   Rule  {22);    Solution 

{23);  Scholium  {24=) 6,  7 

Synopsis 7 

♦-♦-♦ • 

PART  I.-LITERAL  ARITHMETIC. 

OHAPTEE  I. 

FUNDA3IENTAL  BULBS. 

SECTION  I. 

NOTATION. 

System  of  Notation. — Definition  {2S) 8 

Symbols  OF  Quantity.— Arabic  {26);  Literal  {27);  advan- 
tages of  latter  {28) •.  8,  9 

Examples 10,  11 

Laws.— Of  Decimal  Notation  {29);  of  Literal  Notation  {30)  12-11 


X  CONTENTS. 

PAQK 

Symbol  00>  and  its  meaning  {31) 14 

Symbols  of  Opebation.— (3;2) ;  Sign  -f-  (55);  Sign  —  (54); 
Sign—' (55);  Sign  X,  etc.  (36)\  Sign-f,  etc.  (57);  Signy- 

{44) 14-15 

Definitions.— Power  (55)  ;  Eoot  (59)  ;  Exponent  (40),  A 
positive  integer  {41),   A  positive  fraction  [42)^    A  negative 

number  {43)  ;  Examples 15-18 

Symbols  of  Kelation  : — Sign  :  {45^;  Sign  ••  {40)',  Sign  = 

{47) ;  Signs  >  <  {49), 18-19 

Syivibols  of  Aggeegation. —  ,  (),[],  {  },   |,  (50,  51)  19 

Symbols  of  Continuation  {52) 19 

Symbols  of  Deduction  {53) 20 

Positive  and  Negative. — How  applied  {54)  ',  Double  use  of 
signs  -j-  and  —  {55)  ',  Essential  sign  of  a  quantity  {56)  ; 
Illustrations  ;  Abstract  quantities  no  sign  {57)  ',  Less  than 

Zero  {58) ;  Increase  of  a  negative  quantity  {59) 20,  21 

Names  of  Diffekent  Forms  of  Expeession. — Polynomial 
(00)  ;   Monomial,  Binomial,  Trinomial,  etc.  {Gl)  ;  CoeflQ- 

cient  {62)  ;   Similar  Terms  (05) 23,  24 

Exeecises  in  Notation 24,  25 

ExEEcisES  in  Reading  and  Evaluating  Expebssions 25,  26 

Synopsis  and  Test  Questions  26,  27 

SECTION  II. 

ADDITION. 

Definitions. — Addition  {64)  ;  Sum  or  Amount  {65) 28 

Prop.  1. — To  add  similar  terms  {66);  Examples  ;  Cor.  1,  Sign 
of  the  sum  {67)  ;  Sch.  Addition  sometimes  seems  like  Sub- 
traction ;  Cor.  2,  Algebraic  sum  of  a  positive  and  negative 
quantity  {68)  ;  Cor.  3,  Addition  does  not  always  imply  an 
increase  {69)  28-31 

Pkop.  3.— To  ndd  dir.similar  tciraa  (TO);  Sch.  Tlie  sign  — 
before   a  term  ;    Cor.  Adding  a  negative  quantity  (71) ', 

Examples 31-r33 

Peob.  1. — To  add  polynomials  [72, ;  Sck.  1,  Practical  method 
of  adding  ;  Sch.  2,  Literal  and  decimal  addition  compared  ; 

•    Examples 33-37 

Prop.  3. — Terms  but  partially  similar  (73)  ;  Examples 37.  38 

Prop.  4.  —Compound  turms  {74)  ;   Examples  ^ oS.  ."^i) 

Synopsis  and  Te?.t  Qi'estjoks iO 


CONTENTS.  XI 

PAGE 

SECTION  III. 

SUBTKACTION. 

Definitions. — Subtraction,  Minuend,  Subtrahend,  Difference, 
or  Kemainder  (75,  76) 40,  41 

Prob.— To  perform  Subtraction  {77) 42 

Practical  Suggestions,  Sch.  1,  2  ;  Kemoving  a  ( )  preceded 
by  the  sign  —  {78) ;  Introducing  a  ( )  {70} ;  Several  paren- 
theses inclosing  each  other  {80)  45-48 

Examples 42-49 

Synopsis  and  Test  Questions 49 

SECTION  IV. 

MULTIPLICATION. 

Definitions. — {81),    Cors.  1,  2,  and  Sch.  Character  of  the 

factors  and  product  {82,  83,  84) 50 

Peop.  1, — Order  of  factors  immaterial  {83) 51 

Prop.  2. — Sign  of  product  {80)  ;  Cors.  1,  2,  3,  Sign  of  pro- 
duct of  several  factors  {87,  88,  89) 51,  52 

Prop.  3. — Product  of  quantities  affected  with  exponents  (90) ; 

Examples 52-54 

Prob.  1. — To  multiply  monomials  {91)  ;  Examples 54-56 

Prob.  2. — To  multiply  poljTiomials  [92);  Examples 56-59 

Three  Important  Theorems. — Square  of  the  sum  {94); 
Square  of  the  difference  {93);  Product  of  sum  and  differ- 
ence {96 1  ;  Examples 59-61 

Synopsis  and  Test  Questions 62 

SECTION    V. 

DIVISION. 

Definitions,  — {97)  ;  Relation  to  multiplication  (.9^,  99) ; 
6'or.9.  1-5,  Deductions  from  definition  {100-104);  Can- 
cellation {105) 63,  64 

Lemma  1. — Sign  of  quotient  {106) 64 

Lemma  2.  a*"  -^  a"  {107)  ;  Examples  ;  Cor.  1.  Exponent  0 
{108)  ;  Cor.  2,  How  negative  exponents  arise  {109)  ; 
Cor.  3,  To  transfer  a  factor  from  dividend  to  divisor 
{110) 64-66 


XU  CONTENTS. 

PAGS 

Feob.  1. — Division  of  Monomials  {111)  ;  Examples 67,  68 

Pbob.  2. — To  divide  a  polynomial  by  a  monomial  {112);  Ex- 
amples ;  Arrangement  of  terms  {113) 68-70 

Peob.  3. — To  divide  a  polynomial  by  a  polynomial  {lid); 

Examples 70-75 

Synopsis  and  Test  Questions  76 


OHAPTEE  n. 

FACTOItIJ!^G. 

SECTION  I. 

FUNDAMENTAL  PKOPOSITIONS. 

Definitions.— Factor     {115}  ;     Common    Divisor    {116)  ; 

Common     Multiple    (117)  ;     Composite    {118^  ;    Prime 

{119,  120) 77 

Peop.  1. — To  resolve  a  monomial  {121)  ;  Examples 78 

Peop.  2. — To  remove  a  monomial  factor  from  a  polynomial 

{122)  ;  Examples 79 

Peop.  3.— To  factor  a  trinomial  square  {123);  Examples 79-81 

Peop.  4. — To  factor  the  difference  of  two  squares   {124)  ; 

Examples 81,  82 

Peop.  5.  — Given  one  factor  to  find  the  other  {125) ;  Examples  82 

Peop.  6. — By  what  the  sum,  and  the  difference  of  like  powers 

are  divisible  {126)  ;  Examples 83-85 

Frop.  6  applied  to  the  case  of  fractional  exponents  {128)  ; 

Examples 85,  86 

Peop.  7. — To  resolve  a  trinomial  {120)  ;  Examples   86,  81 

Peop.  8. — To  resolve  a  polynomial  by  separating  it  into  parts 

{130);  Examples 87,  88 


SECTION  II 

GKEATEST  OR  HIGHEST  COMMON  DISTESOR. 

Definition  il31) ^8 

Lemma  1.'— The  H.  C.  D.  the  product  of  all  common  factors 

a32)  ;  Examples 89.  90 


CONTENTS.  xiii 

PAQK 

Lemma  2. — A  divisor  of  the  polynomial  of  the  form  Ajc"-}- 

^x"-i4-ac"-2 ^  +  F  {134) 90 

Lemma  3. — ^A  divisor  of   a    number    divides    any  multiple 

{135)  91 

Lemma  4. — A  C,  D.    divides  the  sum,    and   the    difference 

{136) 91 

General  Ruue  for  H.  C.  D.  demonstrated  and  applied  {137 y 

138) 92-98 

SECTION  III. 

LOWEST  OK  LEAST  COMMON  MULTIPLE. 

Pbob. — To  find  the  L.  C.  M.  by  resolving  numbers  into  factors 

{140):  Examples 98,  99 

Finding  the  L.  C.  M.  facilitated  by  the  method  of  H.  C.  D. 

Sch 99,  100 

Synopsis  and  Test  Questions 101 


« • » 

CHAPTEE  III. 

F  B  A  C  T  I  O  N  S. 

SECTION  L 

Definitions   and  Fundamental   Principles.  —  Of  the 

terms  Fraction,  Numerator,  Denominator,  etc,  {14:1).. . .  103 

The  true  conception  of  a  literal  fraction  {142) 103 

Value  op  a  Fraction  {143) 103 

Cor's  1,  3. — Changes  in  terms  of  a  fraction  {144, 145,). .  103 

Uses  and  Definitions  of  various  terms  (146-156) 10S-I05 

Signs  of  a  Fraction  : — 

Three  things  to  consider  {157) 105 

Essential  character  ;  Examples  {158). 105,  lOG 

SECTION  IL 

REDUCTIONS. 

Varieties  of  {159) 107 

Prob.  1.— To  lovrest  terms ;  Examples 107, 109 

Sch.  1.— The  use  of  the  H.  C.  D.  • 108 


XIV  C0JSTENT8. 

PAGE 

ScH.  2. — The  converse  process  ;  Examples 109 

Pbob.  2. — To  improj)er  or  mixed  form  (160);  Examples  109,  110 

Cob. — Use  of  negative  exponents  (161) 110 

Pbob.   3. — From   integral  or  mixed    to  fractional  form 

(162);  Examples Ill,  112 

Pbob.  4. — To  common  denominator  {163)',  Examples. , .  112,  113 

Cob.— To  L.  C.  D.  {164);  Examples 114, 115 

Pbob.  5. — Complex  to  simple  {165);  Examples 115-117 

SECTION  III. 

ADDITION. 

Pbob.— To  add  fractions  {166);  Examples 117-119 

Cob. — To  add  mixed  numbers  {167) ;  Examples 119-121 

SECTION  IK 

SUBTKACTION. 

Pbob. — To  subtract  fractions  {168);   Examples 121-123 

Cob. — To  subtract  mixed  numbers  {169);  Examples 123,  124 

SECTION    V. 

MULTIPLICATION. 

Pbob.  1. — A  fraction  by  an  integer  {170);  Examples 124, 126 

Pbob.  2.  —To  multiply  by  a  fraction  {171);  Examples 126-129 

ScH. — When  no  common  factors 127 

Cos.— To  multiply  mixed  numbers  {172) 129 

SECTION  VL 

DIVISION. 

Pbob.  1.— To  divide  by  an  integer  {173);  Examples 130,  131 

Pbob.  2.— To  divide  by  a  fraction  {174:)\  Examples 131,  136 

Eeason  for  inverting  the  divisor 132 

ScH.  1. — To  reverse  the  operation  of  multiplication 132 

ScH.  2. — Using  the  form  of  a  complex  fraction 135 

REcrPBOCAii,  what  {175) 136 

Synopsis  or  Fbactions  and  Test  Questions 137 


CONTENTS.  XV 

PAGE 

OHAPTEE  IV.     , 

POWERS  AND  ROOTS. 

SECTION  L 

INVOLUTION. 

Note. — The  Fundamental  Principle 138 

Genekai.  Definitions. — Power  (j? 7^);  Root  (177)',   Power 
and  Eoot  correlatives  (178);  Names  of  different  Powers  { 

and  Roots  {179) ;  Exponent  or  Index  ;  How  to  read  an  expo- 
nent {180) ;  Transferring  a  factor  from  numerator  to  denomi- 
nator of  a  Fraction  (181);  Radical  {18*^)',  Radical  Sign 
{183};  Imaginary  Quantity  (J^^j;  'Real  {183);  Similar 
Radicals  {186);  Rationalize  {187);  to  affect  with  an  expo- 
nent {188);  Involution  {189);  Evolution (190) ;  Calculus 

of  Radicals  {191) 138-141 

Involution  : 

Pkob.  1. — To  raise  to  any  Power  {192)  ;  Examples 141-143 

CoK.— Signs  of  Powers  {193)  143 

Peob.  2.— To  affect  with  any  Exponent  {194=) 143 

Examples 144-148 

Pbob.  3.— The  Binomial  Formula  {195) 148 

Applications  148-154 

Form  for  expansion  of  (1  -j-  ^)" 1^0 

Cob.  1. — When  the  series  terminates  (196)  ...  151 

Cob.  2.— Number  of  terms  {197) 151 

Cob.  3.— Equal  coefficients  {198) 152 

Cob.  4. — Sum  of  exponents  in  any  term  {199)  152 

Cob.  5.— Statement  of  the  Rule  {200) 152 

Cob.  6. — Signs  of  the  terms  in  the  expansion  of 

(a  — &)"»  {201) 153 

SECTION  IL 

EVOLUTION. 

Peob.  1. — To  extract  root  of  perfect  power  {202);  Examples  154 

ScH.— Signs  of  root  {203) 155 

Cob.  1. — Roots  of  Monomials  {204:);  Examples 155 

Cob.  2.— Root  of  Product  {205)  156 

Cob.  3.— Root  of  a  Quotient  {206)  ;  Examples 156 


XVI  CONTENTS. 

PAOK 

Prob.  2.— To  extract  the  square  root  of  a  Polynomial,  Eule, 

Dem.  and  Examples  (207) 157-160 

Pbob.  3. — To  extract  the  square  root  of  a  Decimal  Number, 

Kule,  Dem.  (209) ;  Examples  161-164 

CoE. — Eoots  of  Fractions  {210);  Examples 165 

Pbob.  4. — To  extract  the  cube  root  of  a  Polynomial  {211); 

Eule,  Dem.,  Examples 165-169 

Pkob.  5. — To  extract  the  cube  root  of  a  Decimal  Number 

{212);  Eule,  Dem.,  Examples 169-173 

Peob.  6. — To  extract  roots  whose  indices  are  composed  of  the 

factors  2  and  3  {213);  Examples 174 

Peob.  7.  — To  extract  the  mth  root  of  a  Decimal  Number  {214)  1 74 

Peob.  8. — To  extract  the  mth  root  of  a  Polynomial  {215). . .  175 

Examples 175-177 

SECTION  III. 

CALCULUS  OF  EADICALS. 
Reductions  : 

Peob.  1. — To  remove  a  factor  {217) ;  Examples 178-180 

CoR.— To  simpHfy  a  fraction  {218)  ;  Examples 180, 181 

Peob.  2. — When  the  index  is  a  Composite  Number  {210) 

Examples 181, 182 

Prob.  3.— To  any  required  index  {220)  ;  Examples 182, 183 

OoR. — To  put  a  coefficient  under  a  radical  sign  {221) ; 

Examples 183-185 

Peob.  4.— To  a  Common  Index  {222)  ;  Examples 185-187 

Peob.  5. — To  rationalize  a  monomial  denominator  {223)  ; 

Examples 187-189 

Peob.  6. — To  rationalize  a  Eadical  Binomial  denominator 

of  the  2nd  degree  {224);  Examples 189, 190 

Prob.  7. — To  rationalize  any  Binomial    Eadical    {225)  ; 

Examples 190, 191 

Peob.  8.— To  rationalize  Va  -f-  VJT  +  "/c  {226) 192 

SECTION  IV. 

COMBINATIONS  OF  EADICALS. 
Addition  and  Subteaction  : 

Peob.  1.— To  add  or  subtract  {227) ;  Examples 192-194 

MULTIPWCATION  . 

Prop,  I.— Product  of  lil^e  roots  {228) 195 


CONTENTS.  XVU 

PAGE 

Prop.  2.— Similar  Eadicals  (229) 195 

Pbob.  2.— To  multiply  radicals  {230);  Examples 195-198 

DrvisioN  : 

Pbop.— Quotient  of  like  roots  (231) 198 

Peob.  3.— To  divide  radicals  (232);  Examples. 199-201 

Involution  : 

Pbob.  4.— To  raise  to  any  power  (233);  Examples 201,  20i 

Cor. — Index  of  Boot  and  Power  alike  {23d) 202 

Evolution  : 

Pbob.  5. — To  extract  any  root  of  a  Monomial  Kadical 

(;255);  Examples 202-205 

Pbob.  6. — To  extract  the  square  root  of  a  +  n  vT,  or 

m^a    ±  n^b  {237) ;  Examples 205,  206 

Imaginary  Quantities  : 

Definition  {238) ;  not  unreal  {239) ;  a  curious  property 

of  {24:0) 206 

Prop.— Eeduced  to  form  m^—  1,   {241) 207 

Examples  in  Multiplication  and  Division 207,  208 

Stnopsis  and  Test  Questions  on  the  chapter 209,  210 


PART  IL-ALGEBRA. 


OHAPTEE  I. 

SIMPLE  EQUATIONS. 

SECTION  L 

EQUATIONS  WITH  ONE  UNKNOWN  QUANTITY. 

Definitions  :~Equation  {1)  ;  Algebra  {2) ;  Members  (5) ; 
Numerical  Equation  {4) ;  Literal  Equation  {5) ;  Degree  of 
an  Equation  {6) ;  Simple  Equation  (7) ;  Quadratic  Equation 
{8) ;  Cubic  Equation  {9)\  Higher  Equations  {10) 211-213 

Transformations. — What  {11^  12)',  Axioms  {13);  Prdb.  To 

clear  of  fractions  {14);  Examples 213-216 

Illustration  by  balance 214 

Prob.— Transposition  {15,  16);  Examples 216,  217 

Solution  of  Simple  Equations  : — What  {17)  ;When  an  equa- 


XVIU  CONTENTS. 

PAGE 

tion  is  satisfied  (18) ;  Verification  {19);  Proh.  1.  To  solve 

a  Simple  Equation  (20) ;  Examples 218-230 

ScH.  1. — Kinds  of  changes  which  can  be  made  (21) 220 

CoK.  1. — Changing  signs  of  both  members  (22) 223 

ScH.  3. — Not  always  expedient  to  make  the  transformations 

in  the  same  order  (23) 226 

ScH.  4. — Equations  which  become  simple  by  reduction  (J^4)  229 

Simple  Equations  Containing  Kadicals  : — (25) ;    Proh.   2. 

To  free  an  equation  of  Eadicals  (26);  Examples 230-236 

SuMMAKY  or  Pkactical  Suggestions  {27 f  28) 236,  237 

Applications  to  the  solution  of  examples  {29) ;  Statement, 
Solution,  What  {30)',  Knowledge  required  in  making 
statement  {31) ;  Directions  to  guide  in  making  statement 
{32)',  Examples 237-250 

Translations  or  Equations  into  Pkactical  Peoblems  (55) ; 
Examples 250-252 


SECTION   IL 

SIMPLE,  SIMULTANEOUS,  INDEPENDENT  EQUATIONS  WITH 
TWO  UNKNOWN   QUANTITIES. 

Definitions. — Independent  Equations  {37)  ',  Simultaneous 
Equations  {38)  ;  Elimination  {39)  ',  Methods  of  Elimina- 
tion {40) 253,  254 

Elimination. — Peob.  1. — By  Comparison  {4:1)  ;    Examples. .  254-256 

Pkob.  2.— By  Substitution  {42)  ;  Examples 257-259 

Pkob.  3.— By  Addition  or  Subtraction  {43)  ;  Examples  . . .  260-263 

Examples  for  General  Practice 263-265 

Applications 265-271 


SECTION  III 

SIMPLE,  SIMULTANEOUS,  INDEPENDENT  EQUATIONS  WITH 
MOKE  THAN  TWO  UNKNOWN  QUANTITIES. 

Pkob.— To  solve  {46)  ;  Examples 272-277 

Applications 277-279 

Synopsis  and  Test  Questions  on  Chapter  I.,  Part.  II 280 


CONTENTS.  XIX 

FAOE 

OHAPTEE  n. 

RATIO,  JPHOrOMTIOK,  AND  PROGHESSION. 

SECTION  I. 

KATIO. 

Definitions.— Ratio  (47)  ',  Sign  (48)  ;  Cor.  Effect  of  multi- 
plying or  dividing  the  terms  (40)  ;   Direct,  and  Reciprocal 
Ratio  (50,  31 ) ;   Greater,  and  Less  Inequality  {52} ;   Com-    , 
pound  Ratio  {33) ;  Duplicate,  Subduplicate,  etc.  {54) 281-283 

Examples 283-285 

SECTION  II 

PROPORTION. 
Definitions. — Proportion  (55);  Extremes,  and  Means  {56); 
Mean  Proportional  (57)  ;    Third  Proportional   {58)  ;    In- 
version {59) ;  Alternation  {60) ;  Composition  {61) ;  Division 

{62) ;  Continued  Proportion  {64)   285-287 

Peinciples  of  Tbansformation  : 

Peop.   1. — Product   of  extremes   equal    product  of  means 
(65);  Cor.  1.  Square  of  mean  proportional  {66);   Cor.  2. 

Value  of  any  term  {67) 287 

Peop.  2. — To  convert  an  equation  into  a  proportion  {68); 

Examples 288 

Peop.  3. — "What  transformations  may  be  made*  without  de- 
stroying the  proportion  {69) 288 

EXAJMPLES   OF   TeANSFOEMATION  I 

Multiples  of  Terms 289 

Change  in  Order  of  Terms 289-291 

By  Composition  and  Division 291,  292 

Miscellaneous  Exercises 292,  293 

SECTION  III 

rnOGRESSIONS. 

Definitions. — Progression,  Arithmetical,  Geometrical,  As- 
cending, Descending,  Common  Difference,  Ratio  {70)  ; 
Signs  and  Illustrations  {71);  Five  things  considered  {72}..  294,  295 


XX  CONTENTS. 

PAG» 

Akithmetical  Progkession. — Prop.  1.  To  find  the  last  term 
(75)  ;  Prop.  2.  To  find  the  sum  {74)  ;  Cor.  1.  These  two 
formulas  sufficient  (75)  ;  Examples ;  Cor.  1.  To  insert 
means  {76) 295-299 

Geometeical  Peogression. — Prop.  1.  To  find  the  last  term 
{77);  Prop.  2.  To  find  the  sum  {78);   Cor.  1.  These  form- 
ulas sufficient  {70)  ;    Cor.  2.    A  second  formula  for  sum  f 
{80)  ;    Cor.  3.  To  insert  means  {81)  ;    Cor.  4.  Sum  of  an 
infinite  series  {82) ;  Exami^les 299-303 

Synopsis  of  Eatio,  Proportion,  and  Progressions,  and  Test 

Questions 304 

Applications  of  Ratio,  and  Proportion 305-313 


OHAPTEE  EI. 

BUSINESS  RULES   [OF  ARITHMETIC.^ 

SECTION  L 

PERCENTAGE. 
The  Problem  Stated  {83) ;  Prob.  1.  To  express  the  relation 
between  base,  rate  per  cent.,  and  percentage  {84)  ; 
Prob.  2.  To  express  the  relation,  rate  per  cent.,  Amount 
or  Difference,  and.  Base  {85)  ;  Sch.  These  two  formulas 
sufficient ;  Examples 314-316 

SECTION  II 

SIMPLE  INTEREST  AND  COMMON  DISCOUNT. 

pROB.  1. — To  express  the  relation  between  principal,  rate 
per  cent.,  and  time  {86)-  Prob.  2.  To  express  the  rela- 
tion between  amount,  principal,  rate  per  cent. ,  and  time 
{87)  ;  Sch.  These  two  formulas  sufficient ;  Examples ; 
Cor.  To  find  the  time  required  for  principal  to  double, 
triple,  etc.  {89) 317-324 

Prob.  — To  find  one  of  the  equal  periodic  payments  which 
will  discharge  principal  and  interest  {00) ;  Examples 324-326 


CONTENTS.  XXI 

PAGK 

SECTION  III. 

PARTNERSHIP. 

Simple  Paetnekship  {91);  Examples  ;    ^ch.  Rule  deduced. .   327-329 
Compound  Pabtnekship  {92) ;  Examples  ;  Sch.  Rule  deduced  329,  330 

SECTION  IV. 

ALLIGATION. 
Examples  and  Solutions 330  -334 


CHAPTER  lY. 

QUADRATIC   EQUATIONS. 

SECTION  I. 

PURE    QUADRATICS. 

Definitions.  —Quadratic  {94:)  ;  Pure  {96) ;  Affected  {97) ; 
Root  {98) 335 

Resolution  op  a  Puee  Quadeatic  Equation  {99) ;  Cor.  1.  A 
Pure  Quadratic  has  two  roots  {100)  ;  Cor.  2.  Imaginary- 
roots  {101)  ;  Examples 335-340 

Applications 340-344 

SECTION  II 

AFFECTED   QUADRATICS. 

Definition  {102) 344 

Resolution. — Common  method  {103);  Examples 345-355 

Sch.  1.^ — Definition,  completing  the  square 348 

Cob.  1.— Two  Roots,  character  of  {104) 350-351 

Cob.  2. — To  write  the  root  of  x'^  -\-  px  =  q  without  complet- 
ing the  square  {105) 353 

Cob.  3.— Special  methods  {106);  Examples 355,  356 


XXU  CONTENTS. 

PAGE 

SECTION  III. 

EQUATIONS  OF  OTHER  DEGEEES  WHICH  MAY  BE  SOLVED 
AS  QUADRATICS. 

Pbop.  1. — Any  pure  Equation  {107)  \  Examples 357,  358 

Pbop.  2. — Any  equation  containing  one   unknown   quantity 
with  only  two  different  exponents,   one  of  which  is  twice 

the  other  {108) ;  Examples 358-360 

Special  Solutions  ;  Examples  {109-111) 360-366 

SECTION  IV. 

SIMULTANEOUS  EQUATIONS  OF  THE  SECOND  DEGREE 
BETWEEN  TWO  UNKNOWN  QUANTITIES. 

Pbop.  1. — One  equation  of  the  second  degree  and  one  of  the 

first  {112} ;  Examples 367,  368 

Pbop.  2. — Two  equations  of  the  second  degree  usually  involve 

one  of  the  fourth  after  eHminating  {113) 368 

Pbop.  3. — Homogeneous  Equations  {111:)',  Examples 368-370 

Pbop.  4. — "When  the  unknown  quantities   are  similarly  in- 
volved (115) ;  Examples 370-372 

Special  Expedients  {116) ;  Examples 372-374 

Applications 374-379 

Synopsis  of  Quadratics  and  Test  Questions 380 


OHAPTEE  V. 

LOGARITHMS. 

Definition,  Illustrations  and  Examples  {117) 381,  382 

A  system  of  Logarithms,  what  {118) ;  Two  in  use  {119) ...  382 

The  Common  Use  or  Logarithms  {120);  Prop.  1.  Sum  of  log- 
arithms equals  logarithm  of  product  {121) ;  Prop.  2.  Dif- 
ference of  logarithms  equals  logarithm  of  quotient  {122); 
Prop.  3.  Logarithm  of  a  power  {123) ;  Prop.  4.  Logarithm 
of  a  root  {124) 382 


CONTENTS.  XXUi 

FAGB 

Table  of  Logaeithms,  what  {125) ;  Prob.  How  to  make 

{126) 383-385 

Prob.  To  find  the  logarithm  of  a  number  from  a  table 
(127) ;  Characteristic,  Mantissa  {128,  129) ;  Sample  of 

a  Table  ;  Examples 385-389 

Prob. — To  find  a  number  corresponding  to  a  given  loga- 
rithm (150) ;  Examples 389 

Examples  in  the  use  of  logarithms  {ISl) 389, 390 


APPENDIX. 
SECTION    /. 

DIFFERENTIA  TION. 

Definitions 391-394 

Rules  and  Examples 394-401 

Indeterminate  Coefficients 401 

SECTION    II. 
The  Binomial  Formula  demonstrated 402-404 

SECTION    III. 

The  Logarithmic  Series  produced 404  409 

Production  of  a  Table  of  Logarithms 409-411 

SECTION    IV, 
Higher  Equations 411-423 


XXiv  CONTENTS. 

SECTION    V. 

PAGE 

Interpretation  or  Discussion  of  Equations 423-436 

SUCTION    VI. 
Permutations  and  Combinations 436-439 


INTRODUCTION* 

FOR  THE  USE  OF  VERY  YOUNO  PUPILS  WITH  BUT  A 
LIMITED  KNOWLEDGE  OF  ARITHMETIC. 


SECTION  / 
How  Letters  are  used  to  represent  Numbers. 

Exercise  1.  Three  times  5,  and  4  times  5,  and  2  times 
5,  make  liow  many  times  5  ?  3  times  7,  and  4  times  7, 
and  2  times  7,  are  how  many  times  7  ?  3  times  any  nnm- 
ier,  and  4  times  the  same  mimher,  and  2  times  the  same 
mimber,  are  how  many  times  that  number  ? 

Ex.  2.  Two  times  8,  plus  5  times  8,  plus  3  times  8,  plus  1 
time  8,  are  how  many  times  8  ?  2  times  17,  plus  5  times  17, 
plus  3  times  17,  plus  1  time  17,  are  how  many  times  17? 

2  times  a?iy  numher,  plus  5  times  tlie  same  numher,  plus 

3  times  the  same  member,  plus  1  time  the  same  numher,  are 
how  many  times  that  number  ? 

Ex.  3.  Five  23's,  plus  4  23's,  plus  11  23's,  are  how  many 
23's?  ^^is.,  20  23's. 


*  Pupils  who  have  a  fair  knowlecli^e  of  arithmetic,  and  are  somewhat  mature  in 
years,  will  not  need  to  read  this  introduction.  (See  the  second  paragraph  of  the 
preface.)  Little  or  no  attempt  is  made  in  this  introduction,  at  demonstration.  It 
is  desifjned  to  give  the  pupil  simply  some  insight  into  the  character  of  the  literal 
notation,  to  acquaint  him  with  the  mariner  in  which  the  fundamental  operations 
are  perfonned  un  simple  literal  expressions,  and  to  inti'oducc  the  equation  as  a 
method  of  solving  problems.  Of  course  the  teacher  can  introduce  more  or  less 
explanation  of  principles  as  he  may  deem  best  for  the  particular  class.  But  with 
those  for  whom  this  part  ia  prepared,  such  explanations  must  be  necessarily 
largely  oral. 

2 


XXVI  INTRODUCTION. 

Im  In  Algebra  tve  often  use  letters  to  represent,  or  stand 
for,  mmihers.    The  following  exercises  will  show  how : 

Ex.  4.  Suppose  a  stands  for  some  number,  as  in  the  first 
exercise  for  the  5 ;  3  times  a,  and  4  times  a,  and  2  times  a, 
make  how  many  times  a  ?  Again,  suppose  a  stands  for 
some  number,  as  7  in  the  first  exercise ;  3  times  a,  and  4 
times  a,  and  2  times  a,  are  how  many  times  a  ?  Again, 
suppose  a  stands  for  any  numher,  only  that  it  shall  mean 
the  same  niiniber  each  time ;  3  times  a,  and  4  times  a,  and 
2  times  a,  make  how  many  times  a  ? 

Ex.  5.  If  m  stands  for  (represents)  some  number;  how 
many  times  ni,  are  2  times  m,  plus  5  times  m,  plus  3  times 
m,  plus  1  time  m  ?  Does  it  make  any  difference  what 
number  m  stands  for,  so  that  it  means  the  same  number  all 
the  time  ?     Compare  this  with  Ex.  2. 

Ex.  6.  Suppose  b  represents  some  number  (meaning  the 
same  number  all  the  time  in  this  exercise),  5  b's,  plus  4  Z»'s, 
plus  11  b^s,  are  how  many  b's  ?     Compare  with  Ex.  3. 

2.  Thus  we  7nay  use  any  letter  to  represent  any  number, 
only  so  that  it  alivays  means  the  sa^ne  number  in  the  same 
exercise. 

3.  Wien  a  letter  is  used  to  represent  a  number,  the  figure 
which  tells  hoio  many  times  the  mwiber  represented  by  the 
letter  is  taken,  is  just  loritten  before  the  letter,  the  luord 
"  times "  being  left  out.  Thus  3«  means  3  times  a,  ib 
means  4  tiiiies  b,  ^m  means  7  times  m,  105a;  means  105 
times  the  number  represented  by  x,  whatever  that  number 
may  be. 

X  4.  The  number  placed  before  a  letter  to  tell  hoio  many 
times  the  letter  is  taken,  is  called  a  Coefficient.  If  no 
figure  stand  before  a  letter,  the  letter  is  taken  once,  or  its 
coefiicient  is  said  to  be  1.    Thus,  m  means  one  time  m. 


HOW  LETTERS   REPRESENT  NUMBERS.  XXVll 

Ex.  7.  How  many  times  the  number  represented  by  h, 
are  4J,  3J,  6^,  and  h  ?  That  is,  4  times  some  number,  plus 
3  times  the  smne  numher,  plus  6  times  the  same  numher, 
plus  one  time  the  same  numher,  are  how  many  times  that 
number  ? 

Ex.  8.  ba,  plus  a,  plus  6tt,  plus  8a,  are  how  many  times 
a  ?  ^?Z5.,  20«. 

Query. — If  the  a  in  ^a  meant  one  number,  the  a  alone  another 
number,  the  a's  in  6«  and  la  still  other  numbers,  could  you  answer 
this  exercise  in  the  same  way  ?  You  could  not  answer  it  at  all. 
The  a  must  mean  the  same  number  all  the  time,  in  the  same  ex- 
ample. 

Ex.  9.  10a  +  5«  +  la  +  2a,  *  are  how  many  times  a  ? 

Query.— Is  it  necessary  that  a  should  mean  the  same  thing  in  this 
exercise  that  it  did  in  Ex.  8  ? 

Ex.  10.  Za  +  2a  +  ba  +  8a,  are  how  many  times  a? 
Ans.,  18a.  How  much  is  this  if  a  is  G  ?  Ans,,  108.  How 
much  if  a  is  11  ?     Ans.,  198. 

Ex.  11.  Eleven  times  8,  minus  5  times  8,  are  how  many 
times  8  ? 

Ex.  12.  11a  —  ba  are  how  many  times  a  ?  Eleven  times 
any  number,  minus  5  times  the  same  number,  are  how 
many  times  that  number? 

Ex.  13.  12a;  —  Ix  =  how  many  times  x  ?  How  much 
is  this  if  X  represents  3  ?  If  a;  represents  2-J-  ?  A7is.  to 
the  last,  11. 

Ex.  14.  bh  +  4:b  +  10b  —  125  =  how  many  times  b  ? 

Ex.  15.  How  much  is  dm  +  ^n  —  ^m  +  6m  —  bm  — 
2m? 


*  The  pupil  19  presumed  to  be  acquainted  with  the  use  of  the  signs.    They  are 
•xplaiued  m  the  body  of  the  full  treatise,  simply  as  a  part  of  the  science. 


XXVm  INTRODUCTION. 

We  do  thi3  thus :  Sm  +  8m  are  11m,  ll?7i  —  4m  are  7m,  7m  +  6m 
are  1dm,  13m  —  5m  are  Sm,  8m  —  2m  are  6m.  Hence  the  answer  is 
6m. 

S.  In  such  examples  as  the  last  you  can  add  together  all  the 
quantities  with  the  +  sign  into  one  sum,  making  in  this  case 
17m,  and  all  those  with  the  —  sign  into  another,  making 
in  this  case  Ihriy  and  then  subtract.  Thus,  17^^  —  ll7)i 
is  Qnif  the  same  result  as  before.  This  is,  generally,  the 
better  way  to  do  such  examples. 

Ex.  16.  How  much  is  lOx  —  4:X  —  ^x  +  dx  —  Sx  +  llo;? 
How  much  is  it  if  x  is  3  ? 

Ex.  17.  How  much  is  10:r  —  15:?:  ? 

Sug's. — Of  course  we  cannot  take  15a?  out  of  10a;.  But  we  can 
take  10.C  of  the  15x  from  the  first  10a;,  and  there  will  then  remain  5x 
of  the  15a;,  which  cannot  be  taken  out  of  the  10a;.  We  indicate  this 
by  writing  it  —  5a;.  This  means  that  5a;  was  to  be  subtracted,  but 
that  we  liad  nothing  to  take  it  from. 

Ex.  18.  How  much  is  Sx  —  5x  —  2x?         Ans.f  —  4a;. 

Query. — What  does  this  answer  mean  ? 

Ex.  19.  How  much  is  12a  +  3a  —  6a  -  20a  ? 


Ex.  20.  How  much  is  2  times  3  times  a  certain  number, 
as  5  ?  A71S.,  6  times  the  number. 

Ex.  21.  How  much  is  5  times  7m  ?  that  is,  5  times  7 
times  a  number  which  we  will  represent  by  m  ? 

Ans.,  35  times  w?,  or  35m. 

Ex.22.  How  much  is  6  times  Sa?  7  times  3a?  10 
times  7b?  9  times  8?/  ?  9  times  8  times  a  number  are 
how  many  times  that  number? 

Ex.  23.  How  much  is  3  times  7m,  and  4  times  Sm,  minus 
2  times  lOwi,  if  m  represents  6  ? 


HOW   LETTERS   REPRESENT   NUMBERS.  xxix 

Ex.  24.  What  is  10a  divided  by  2  ;  that  is,  what  is  J  of 
10a  ?    27x  divided  by  9  is  how  much  ? 

A71S.  to  the  last,  Sx. 

Ex.  25.  How  many  times  a  number  is  10  times  that 
number,  divided  by  2  ;  that  is,  J  of  10  times  a  number  ? 

Ex.  26,  How  much  is  ^  of  48.r  ?  25a;  divided  by  5  ?  y\ 
of  11a;  ?     11a;  divided  by  11  ?     7x  divided  by  7  ? 

Ex.  27.  Divide  10a;  by  5,  then  add  Sx,  then  multiply  by 
2,  then  subtract  4a;,  then  divide  by  3.    What  is  the  result  ? 

[Note.— The  teacher  should  extend  such  exercises  until  his  pupils 
can  perform  them  mentally,  as  rapidly  as  one  would  naturally  pro- 
nounce them.] 


SECTION  IL 
More  about  representing  Numbers  by  Letters. 

X  6.  Wlien  two  letters  refresenting  numlers  are  written 
side  lij  side,  as  in  a  word,  their  product  is  indicated. 
Thus,  ah  means  the  product  of  the  two  numbers  repre- 
sented by  a  and  h,  3abc  means  3  time^  the  product  of  the 
numbers  represented  by  a,  I,  and  c. 

[Note. — Instead  of  saying,  as  above,  the  number  represented  by  a, 
we  usually  simply  say  "  the  number  «,"  or,  "  a,"  without  using  the 
word  number  at  all.  Thus  we  say  3  times  the  product  of  a,  6, 
and  c] 

7.  If  we  want  to  rejjresent  the  jJroduct  of  a  numher  rep- 
resented hy  a  letter,  as  a,  hy  itself  a  certain  numher  of 
times,  instead  ofivriting  aa,  or  aaa,  etc.,  as  we  might,  we  write 
a^  a^  etc.  Thus  J*  means  the  same  as  dhlh.  a^  is  read  "  a 
square  ;  "  a\  "a  cube  ; "  i\  "  i  fourth  power ;  "  x\  "  x  fifth 
power,"  etc. 


XXX  INTRODUCTION. 

^'The  little  figure  iiilaced  at  the  right  and  a  little  above  the 
letter  is  one  form  of  what  is  called  an  Exponent ;]  hut 
theiminl  must  not  get  the  idea  that  all  exjjonents  meaiijust 
what  has  now  been  explained.  This  is  the  case  only  ivhen 
the  ex])onent  is  a  whole  number  loitliout  any  sign  or  loith 

2 

the  +  sign.  Thus,  a~^  does  not  mean  aaa.  Nor  does  a^ 
mean  any  such  thing,  although  the  -  3,  and  the  f  are  expo- 
nents.   What  these  do  mean  will  be  explained  in  due  time. 

[Note. — It  is  of  the  utmost  importance  that  the  pupil  be  guarded, 
from  the  outset,  against  the  notion  that  an  exponent  necessarily  in- 
dicates a  power.  This  false  notion  once  in  the  head,  plagues  the 
pupil  ever  after.] 

Ex.  1.  How  much  is  \a^b,  \i  a  =2,  and  5  =  5?  How 
much  Za^Wx,  if  a  =  3,  Z>  =  2,  a;  =  8  ?  How  much  \W&y\ 
if  «  =  2,  c  =  1,  ^  =  3  ? 

Ex.  2.  How  much  is  a^¥  +  2«?y  —  %,  if  a  =  4,  5  =  3, 
^  =  2  ?  How  much  'Za^bif  —  lay""  +  o5,  the  letters  having 
the  same  values  as  before.    How  much  hby  —  'Zab^  +  4a^3/'' 

-2a? 

Ex.  3.  How  many  times  ab""  is  4«&'*  +  2a5'  —  Zab^  ? 
How  many  times  a^y  is  lOa^y  +  4«'?/  —  Oa^y  —  a'^y  ? 

Ex.  4.  How  man*y  times  ani^y^  is  4  times  Sant'y^  ?  How 
much*  is  6  times  2anfy^  ?  4  times  7a'^b'^c''  ?  10  times 
ISimfx'  ? 

Ex.  5.  How  many  times  ax  is  |  of  20ax  ?  |  of  S5ax  ? 
102ax  divided  by  3  ?  How  much  is  ^  of  72a'x'  ?  125a;y 
divided  by  25  ?     ISmf  divided  by  9  ? 

8.  We  have  learned  in  arithmetic  that  representing 
numbers  by  the  figures  1,  2,  3,  4,  etc.,  is  called  Arabic  or 
Decimal  Notation.    In  like  manner,  representing  numbers 

♦  This  mcaus  the  same  as  the  preccdiujj  qucstiou, 


HOW  LETTERS  KEPKESENT  NUMBERS.  xxxi 

by  the  small  letters  of  the  alphabet,  as  a,  h,  c,  dy x, 

y,  etc.,  is  called  Literal  Notation.  The  pupil  will  see  that 
this  Literal  Notation  is  altogether  a  different  thing  from 
the  Roman  Notation,  in  which  the  seven  capital  letters,  I,  V, 
X,  L,  C,  D,  M  are  used. 

Because  the  Literal  Notation  is  so  much  used  in  Alge- 
bra, it  is  often  called  the  Algebraic  Notation.  But  this 
notation  is  just  as  much  used  in  some  other  branches  of 
Mathematics,  as  in  Algebra. 

0.  An  expression  like  Ha^x,  without  any  other  joined 
with  it  by  the  signs  +  or  — ,  is  called  a  Term,  or  a 
monomial.  If  there  are  tivo  such  terms  joined  together 
by  either  of  the  signs  +  or  — ,  the  two  taken  together  are 
called  a  binomial,  as  6JV  +  2«?/^,  or  10a;  —  Say.  If 
three  terms  are  joined  in  this  way  it  is  called  a  Trino- 
mial,  as  Sa^y  —  2ah  +  21a\  Any  expression  consisting 
of  more  than  one  term  is,  in  general,  called  a  ^olyno- 
7nial. 

Ex.  6.  Point  out  the  monomials,  binomials,  trinomials, 
and  polynomials  in  the  following :  '^ax  —  3^^  bxy  —  6cd  -f 
a  —  2y,  Sa'^nfxy,  &  —  <:r,  a  +  m,  a  +  b  +  c  —  d, 
22oaWc'^d^,  abed,  a  —  b,  ab,  c  —  x^y  +  ax,  x'  +  ?/% 
10^'  +  Sxy. 

10.  Terms  Avhich  have  the  same  letters,  affected  with  the 
same  exponents,  are  called  Similar,  Thus  12a^y,  Qa^y, 
and  —  Sa'^y  are  similar;  but  12ay,  ^a^y,  Sex,  are  not 
similar. 

Ex.  7.  Point  out  the  similar  terms  among  the  folloAving: 
Sax%  2ax,  -  5a'x\  ax,  17ax\  16cy\  -  12cY,  Sa\v% 
—  bcy^,  Qcy'',  l^ax",  c^y". 

11.  Terms  having  the  +  sign  are  called  Positive^ 
and  those  having  the  —  sign,  Negative, 


XXXll  INTRODUCTION. 

SECTION  III, 

How  Numbers  are  Added  in  tlie  Literal  Notation. 

12 »  RULE. — Write  the  expressions  so  that  simi- 
lar TERMS  SHALL  FALL  IN  THE  SAME  COLUMNS.  COMBINE 
EACH  GROUP  OF  SIMILAR  TERMS  INTO  ONE  TERM,  AND 
WRITE  THE  RESULT  UNDERNEATH  WITH  ITS  OWN  SIGN". 
The    POLYNOMIAL   THUS   FOUND   IS   THE    SUM    SOUGHT. 

Ex.  1.  Add  hax  —  Icy,  3ax  +  ^cy,  cy  —  2ax,  —  4:ax  — 
dcy,  —  ax  +  hey,  and  2ax  +  2cy. 

Operation.  * 
Having  written  the  numbers  so  that  similar  terms         ^ax  —  2c7/ 
fall  in  the  same  column,  we  may  begin  to  add  with         dax  +  4cy 

any  column  we  choose.     Adding  the   right-hand  —  2ax  +    cy\ 

column    we   find  it  makes  +  Icy,   and  write  this  —  ^ax  —  3cy 

sum  underneath  the  column  added.    In  like  man-  —    ax  +  5cy 
ner  the  other  column  makes  Zax  (or  +  ^ax),  which  2ax  +  2cy 

we  write  without  any  sign,  as  +  will  then  be  un- — 

derstood.    The  sum  is  Zax  +  ley.  dax  +  Icy 

Ex.  2.  Ex.  3. 


bed  -    2a  + 

4cxy 

2cd  +  3rt  - 

bxy 

8a - 

2xy 

Qcd              + 

14:xy 

-    Za  — 

Ixy 

+  11a  + 

xy 

^cd  -  Iha 

'  10a77i  -    Mif  +    2a^x 

—  Qam  +  4^dy^  —  lOiO^x 
Aam  —  Sdy""  —  2a''x 
lam  -  ISdy'  +     Qa'x 

—  9am  +    2dy^  —    ha'x 

am  +  ISf/i/'  —      a'x 


bed  4-    2a  +     bxy  lam  —  lOa'x 

*  It  is  not  proposed  to  attempt  demonstrations  in  this  introduction,  but  merely 
to  acquaint  the  student  with  a  few  of  the  more  elementary  facts  and  processes  of 
the  science.  Judicious  teachers  will  give  more  or  less  oral  explanation,  accordiptf 
to  the  character  of  the  class. 

t  When  no  sign  is  expressed  before  a  term,  +  is  understood. 


SUBTRACTION   IN   THE  LITERAL   NOTATION.      XXXIU 

Ex.  4.  Add  5rc  -  3a  +  5  +  7  and  -  4a  -  3a;  +  2^>  -  9. 

Sam,  2x  -7a  -\- 3b  —2. 

Ex.  5.  Add  2a  +  35  -  4c  -  9  and  5a  -  3b  +  2c  -  10. 

Ex.  6.  Add  3a  +  2b  —  5,  a  +  5b  —  c,  and  6a  —  2c  +  3. 

Ex.  7.  Add  6xy  -  12a;',  -  4:x'  +  3xij,  4a;'  -  2xy,  and  -• 
3xy  +  4a;'.  Sum,  4:xy  —  Sx'. 

Ex.  8.  Add  3a'  +  Uc  -  e'  +  10,  -  5a'  +  6bc  +  2^'  - 
15,  and  -  4a'  -  9bc  -  lOe'  +  21. 


SECTION  IV, 

How  Numbers  are  Subtracted  in  the  Literal  Notation, 
13.  BULK— Cb.a2^gb  the  signs  of  the  terms  in 

THE  SUBTRAHEND  FROM  +  TO  — ,  OR  FROM  —  TO  +, 
OR  CONCEIVE  THEM  TO  BE  CHANGED,  AND  ADD  THE  RE- 
SULT TO  THE   MINUEND. 

Ex.  1.     From  5bij   -   6a'    +    3a;'  subtract  2by  +  3a' 
+  x\ 

Operation. 

5by  —  Qa^  +  3a;'     Minuend. 
—  2by  —  3(Z*  —    X*     Subtrdliend  with  signs  changed. 

The  Bemainder  sought,  which  is  the  sum  of  the 


—  ^a^  +  2x^  \  minuend  and  the  subtrahend  with  its   signs 
changed.* 

Ex.  2.  From  3aa;    +  51"' y^  —  2m'  take  3aa;  —  b'^y^  —  3m'. 

Bern.,  eby  +  m\    . 

*  If  the  teacher  thinks  it  best,  he  can  give  a  familiar  explanation  of  the  prin- 
ciple involved  ;  yiich  as  is  foiincl  on  pages  41,  42,  43,  of  Part  I.  If  any  explana- 
tion is  given,  it  will  be  well  to  have  it  the  same  in  substance  as  found  in  the  body 
«)f  the  work,  for  obvioua  reasons. 

2* 


XXXIV  INTRODUCTION. 

Ex.  3.  From  10a  —  3b  -h  2c  —  x'  subtract  b  —  5c  + 
x\  Rem.,  10a  —  4&  +  7c  -  2x\ 

Ex.  4.  From  I2xy  —  3c'  +  ah  take  Qxy  +  c'  —  2aZ>. 

Ex.  5.  From  bc'y  —  oal)  take  m,x  —  'Hc'y. 

Rem.,  '^&y  —  Sab  —  mx. 

Ex.  6.  From  x^  —  lla:?/^  +  3«  take  —  6xyz  -{•  7  —  2a 
—  5xyz.  Rem.,  x^  -\-  5a  —  7. 

14.  When  there  is  a  term  in  the  subtrahend  which  has 
no  similar  term  in  the  minuend,  we  see  that  this  term  ap- 
pears in  the  remainder  with  its  sign  changed. 

Ex.  7.  From  6«^'  —2by'  +  4:X  -  3cy  take  -  2ac'  + 
3b^y  —  3x  —  3cy  +  m. 

Rem.,  Sac"  —  2by^  —  3b^y  +  7x  —  m, 

Ex.  8.  From  Sm'x^  -3a  — U  take  a'  -  b\ 

Rem.,  %m\c'  -  3a  -  ^h  -  «'  +  b\ 

Ex.  9*  From  a"  +  2ah  +  b^  take  a^  -  2ab  +  b\ 

Ex.  10.  From  «'  +  b^  take  «'  -  b\ 

Ex.  11.  From  3cm  -y  take  2b  -  3c. 

Ex.  12.  From  x^  —  2xy  +  1  take  ixy. 


SECTION    V. 
How  Numbers  are  Multiplied  in  the  Literal  Notation. 

15.  R  ULE. — To  MULTIPLY  TWO  MONOMIALS  TOGETHER, 
MULTIPLY  THE  NUMERICAL  COEFFICIENTS  AS  IN  THE 
DECIMAL  NOTATION,  AND  TO  THIS  PRODUCT  AFFIX  THE 
LETTERS  OF  THE  FACTORS,  AFFECTING  EACH  LETTER  WITH 
AN  EXPONENT  EQUAL  TO  THE  SUM  OF  THE  EXPONENTS 
OF  THAT   LETTER   IN   BOTH    THE   FACTORS.      If    THE   SIGNS 


MULTIPLICATION   IN   THE   LITERAL   NOTATION.     XXXV 

OF  BOTH  THE  FACTORS  ARE  ALIKE,  THE  PRODUCT  MUST 
HAVE  THE  +  SIGN  ;  BUT  IF  THE  SIGNS  OF  THE  FACTORS 
ARE  UNLIKE,   THE   SIGN   OF  THE   PRODUCT   MUST   BE  — . 

Ex.  1.  Multiply  6ax'  by  3a'x\ 

Operation. — If  we  wish  we  may  write  the  factors  as  ia 
arithmetic,  though  this  is  not  necessary.    Tlie  product  of       Sax"^ 
the  numerical  coefficients  3  and  5  is  15.    To  tliis  affixing      da^x^ 

the  letters  a  and  x,  and  as  a  has  an  exponent  1*  in  the  mul-  

tiplicand,  and  3  in  the  multiplier,  giving  it  an  exponent  3    loa^x^ 
in  the  product,  and  x  an  exponent  5  for  a  like  reason,  we 
have  15a^x^,  as  the  product  of  5ax'^  x  dtC^x^. 

Ex.  2.  Multiply  10  ^nn'  by  3^^^';^'.  3xy  by  4:x'y\  7cx  by 
Sex.    2a  by  a^. 

Ex.  3.  Multiply  -  5a'  by  6b.  Product,  -  SOa'h  - 

Ex.  4.  Multiply  -  Sa'x  by  —  2a'x'i/. 

Product,  (ja^x^y. 

Ex.  5.  Find  the  products  of  the  following :  —  lla'x  by 
2axy ;  X^c^mx^  by  —  dvfx ;  9a  by  4J ;  —  am  by  —  xy ;  ^c^dj^ 
hj  —  ad]  —  bxy  by  —  x'^y'^. 


16.  To  imdtij^ly  tivo  factors  together  ichon  one  or  loth 
are  2)olynomials. 

R  TILE. — Multiply  each  term  of  the  multiplicand 

BY  EACH   TERM   OF   THE   MULTIPLIER,  AND  ADD   THE  PROD- 
UCTS. 

Ex.  1.  Multiply  2a^^•  -  Uy  +  A.m  by  2rt'^'m. 

Operation.  —  It    is    immaterial  2«2^  —  Zhy  +  4?7^ 

where  we    write    the    multiplier,  "la^V^m 

but  we  may  as  well  write  it  as  in 

arithmetic.      So  also  it  makes  no  ^w'lfimx  —  Qd^U^my  +  Qa^h 
difference  whether  we  begin  at  tlie 


When  uo  exponent  is  expressed,  1  is  always  understood. 


37.0„i2 


XXXVl  INTRODUCTION. 

right  hand  or  the  left  to  multiply.  It  is  customary  to  arrange 
the  letters  in  a  term  alphabetically-^  thus  we  write  ^a^bhnx,  instead  of 
^¥a^X7n,  or  au}'-  such  form.  There  would  be  really  no  difference  in 
the  value  of  tlie  terms,  however,  in  whatever  order  the  letters 
came. 

i 
Ex.  2.  Multiply  Qax  —  5a'x  +  Sax'  by  2«V ;   2my  - 

dcy'  by  5m^c' ;  4«5  —  Scd  +  x  hy  ay,  a  +  h  —  c  by  a; ; 

lOa^Di^x  —  4:any  +  Smx'  by  (dO^y^. 

Ex.  3.  Multiply  Wx"  -  2aif  +  bxy  by  a"  -  2xy. 

Operation.—  da^x^  —  2ay^  +  5xy 

a^  —  2xy 


Product  by  a\               da'x^  -  2aY  +  5«^iry 
Product  by  —  2xy,     —  Qa^x^y  +  4:axy*  —  lOx'^y'^ 
Sum  of  partial 


products,  da*x'^  —  2a^y^  +  5a'^xy  —  Qa'^x^y  +  Aaxy*  —  lOx'^y* 

There  being  no  similar  terms  in  these  partial  products,  we  can 
add  them  only  by  connecting  them  Avith  their  proper  signs. 

Ex.  4.  Multiply  x""  —  2xy  +  ?/'  by  2x  —  dy. 

Operation.—  x"^  —  2xy  +  y"^ 

2x-Sy 


Product  by  2x,  2x^  -  Ax'y  +  2xy^ 

Pj-oduct  by  -  3y,  -  Zx'^y  +  Qxy^  ■ 


Entire  product,  2x''  -  Wy  +  ar^^  -  3y^ 

Ex.  5.  Multiply  a+  Jihja  —  n.  Prod,  a^ 

Ex.  G.  Multiply  «*  +  c^'  +  c^"  +  «  +  1  by  a'  —  1. 
Ex.  7.  Multiply  «'  -  ^al  +  V  by  a'  +  ^ah  +  h\ 
Ex.  8.  Multiply  m  +  n  by  m  +  n, 
Ex.  9.  Multiply  m  —  n  by  m  —  n, 
Ex.  10.  Multiply  4a;'  -  9^'  \)y  x  +  2y, 


DIVISION   IN   THE  LITERAL  NOTATION.  XXXVll 

Ex.  11.  Multiply  together  x  —  S,  x  —  5,  a;  —  4,  and  x 
-7. 

-  Ex.  12.  Multiply  x^  -{•  y""  +  z^  —  xy  —  xz  —  yzhjx  -{-  y 
+  z.  Prod,  x^  +  y^  +  z^  -  ^xyz. 


SECTION  VL 

How  Numbers  are  Divided  in  the  Literal  Notation, 

17.  To  divide  when  dividend  and  divisor  consist  of  the 
same  quantity  affected  with  ex])onents. 

RULE. — The   quotient   is    the   common   quantity 

AFFECTED  WITH  AN  EXPONENT  EQUAL  TO  THE  EXPONENT 
OF  THE  DIVIDEND  MINUS  THE  EXPONENT  OF  THE  DIVI- 
SOR.   If  the  signs  of  the  divisor  and  dividend  are 

ALIKE,  THE  SIGN   OF  THE   QUOTIENT  IS  +  ;   IF  UNLIKE,   — . 

Ex.  1.  Divide  a'  by  a;.  Quot.,  a\ 

The  student  will  observe  that  the  product  of  the  divisor  and  quo- 
tient must  always  equal  the  dividend.    In  this  case,  a^  x  o?  =.  a^ 

Ex.  2.  Divide  -  x'  by  x\  Quot.,  -  x\ 

Ex.  3.  Divide  —  m^  by  —  m^.  Quot.,  m. 

Ex.  4.  Divide  ¥  by  -  1)\  Quot.,  -  h\ 

Ex.  5.  Give  the  quotients  in  the  following  cases  •  if  -^  y* ; 
-  x''  -i-  x';  a'  -^  -a';  -  c'  -^  -  c\ 


18.  To  divide  one  monomial  hy  anot^ier, 

RULE. — Divide  the  numerical  coefficient  of  the 
dividend  by  the  numerical  coefficient  of  the  divi- 
sor, AND  TO  THE  QUOTIENT  ANNEX  THE  LITERAL  FAC- 
TORS,   AFFECTING   EACH   WITH    AN    EXPONENT    EQUAL  <  TO 


XXXVlll  INTRODUCTION. 

ITS  EXPOiq-EN^T  1^  THE  DIVIDEND  MINUS  THAT  IN  THE 
DIVISOR,  AND  SUPPRESSING  ALL  FACTORS  WHOSE  EX- 
PONENTS ARE  0.     The  sign  of  the  quotient  will  be 

+  WHEN  THE  DIVIDEND  AND  DIVISOR  HAVE  LIKE  SIGNS, 
AND  —  WHEN  THEY  HAVE  UNLIKE   SIGNS. 

Ex.  1.  Divide  Ub'x'y  by  Sbxij.  QuoL,  hVx, 

Ex.  2.  Divide  %\a\i^mf  by  -  Wmnij. 

Qtiot.,  —  dm^y. 
Ex.  3.  Divide  -  lOoa'y'  by  -  21«^  Quot,  baif. 

Ex.  4.  Divide  —lS7nn''x  by  Qmnx.  Quot.^  —  Zn, 

Ex's  5  to  9.  Give  the  quotients  in  the  following : 
12«y  -^  3rt/;  -  64ay  -^  16.?y ;  ^Ix'xf  -^   -  mx'f-. 


10.  To  divide  one  monomial  hy  another  tuhen  the  coeffi- 
cient in  the  dividend  is  not  divisible  hy  that  in  the  divisor, 
or  the  exponent  of  any  letter  is  greater  in  the  divisor  than 
in  the  dividend,  or  there  is  a  letter  in  the  divisor  not  found  in 
the  dividend. 

RULE. — Write  the   divisor   under  the  dividend 

IN  THE  FORM  OF  A  COMMON  FRACTION,  AND  THEN  RE- 
DUCE THIS  FRACTION  TO  ITS  LOWEST  TERMS  BY  CAN- 
CELLING ALL  FACTORS  COMMON  TO  BOTH  NUMERATOR 
AND   DENOMINATOR.* 

The  sign  of  the  quotient  is  determined  as  in  the  last 
case. 

Ex.  10.  Divide  l^a'x'  by  4«V. 

Operation,    l^a^x^  -r-  A.a^x'^  =  -7-,— r-    Now  iu  18  and  4  there  is 
a  common  factor  2  which  can  be  cancelled ;  the  a^  in  the  numerator, 
♦  The  pupil  is  presumed  to  be  familiar  with  this  operation  iu  arithmetic. 


DIVISION   IN  THE   LITERAL  NOTATION.  ^XXIX 

which  is  two  factors  of  a,  will  cancel  two  factors  of  a  from  a'  in 
the  denominator,  and  leave  a  factor  a  in  the  denominator,  and  in 
like  manner  x"^  in  the  denominator  cancels  x'^  from  the  x^  in  the 

numerator,  leaving  x  therein.    Hence,  ,  „  „  =  — ,  and  —  is  the  re- 

quired  quotient 

Ex.  11.  Divide  12wy  by  lOm'y. 

12jnY       6^2 


Operation.    \2m'^y^  -r-  IQm^y  = 


\Onfy       5m 


3y 


Ex.  13.  Divide  iSaVy'  by  -  32a'xy.      Quot.,  -  \  . 
Ex.  13.  Divide  -  ^Wx'y  by  -  40^>V.  Qiiot,  |^. 

Ex.  14.  Divide  Qax  by  Ihy.  Quot,  ^, 

Ex.  15.  Divide  —  Wmy  by  \^nx.  Quot.,  —  tt.— . 

Ex's  15  to  20.  Find  the  quotients  in  the  following  cases : 
24a*a;'^  -h  Ibmx^ ;  —  "Hax^  by  —  14«??i* ; '  —  bab  by  bxy ; 
IQa^y""  by  —  Vla'y'^ ;  7aV  by  —  Sx^ ;  —  imn^  by  Smn. 

20.  To  divide  a  polynomial  by  a  monomial, 

RULE. — Divide   each  term    of    the    polyn-omial 

DIVIDEND  BY  THE-  MONOMIAL    DIVISOR  ;    AND   WRITE    THE 
RESULTS  IN  CONNECTION  WITH  THEIR  OWN  SIGNS. 

Ex.  21.  Divide  W'x'y''  -  Ua'xy  +  Iba'xy  by  SaVy'. 
Operation.—        3a^.cV)6a^.i;V'  -  12a'a;\iy°  +  15a*x^y' 
2xY  -  4.oxy*  +  5a^x^y 

Ex.  22.  Divide  12a*^'  -  IQaY  +   20«y  -  2Say  by 
-  4:ay. 
Ex.  23.  Divide  loa'bc  -  20acy^  +  6cd^  by  -  babe, 

Quot,-^a  +  -^~-^. 


INTRODUCTION. 

Ex.  24.  Divide  IGwiV  -  2^m'x''  +  Wm'x'  by  2mV. 
Ex.  25.  Divide  35«'%'  +  2Sa'bY  -  4:9ai'y*  by-  laby. 


21.  To  perform  division  lolien  both  dividend  and  divisor 
are  ^polynomials, 

BULK — 1st.  Arrange  both  dividend  and  divisor 

WITH   REFERENCE  TO   SOME   LETTER    FOUND  IN   BOTH,   i.e., 

place  that  term  first  at  the  left  hand  which  has  the  highest 
exponent  of  this  letter,  the  term   containing  the  next 
highest  exponent  of  this  letter  next,  etc. 
2nd.  Having  arranged  dividend  and  divisor  thus, 

DIVIDE  the  first  TERM  OF  THE  DIVIDEND  BY  THE  FIRST 
TERM  OF  THE  DIVISOR  FOR  THE  FIRST  TERM  OF  THE  QUO- 
TIENT ;  THEN  SUBTRACT  FROM  THE  DIVIDEND  THE  PRODUCT 
OF  THE  DIVISOR  INTO  THIS  TERM  OF  THE  QUOTIENT,  AND 
BRING  DOWN  AS  MANY  TERMS  TO  THE  REMAINDER  AS  MAY 
BE  NECESSARY  TO  FORM  A  NEW  DIVIDEND.  DiVIDE  AS 
BEFORE,  AND  CONTINUE  THE  PROCESS  TILL  THE  WORK  IS 
COMPETE. 

Ex.  1.  Divide  Ga'^x'  -  4:ax'  -  4:a'x  +  x'  -\- a'  hy  —  2ax 

+  «*  +  x\ 

Operation.— These  polynomials,  arranged  according  to  the  letter 
a,  and  placed  in  the  ordinary  manner  f(H-  division,  become 

a^  —2ax  +  x'^)a*  -  A^a^x  +  Qa'^x''  —  4ax^  +  x\ar'  —  2ax  +  x^ 
a*  -  2a^x  +    a'x^ 

—  2a^x  +  5a\i^^  —  Aax^ 

—  2a?x  +  4«-.f^  —  ^ax^ 

a'\v^  —  2«.c^  +  X* 
n-X'  —  2ax^  +  x* 

The  pupil  will  have  no  difficulty  in  following  the  work,  if  he  com- 
pares it  carefully  with  the  rule.  The  polynomials  might  equally 
well  have  been  arranged  with  reference  to  x.  Thus  x'^  —  2ax  +  a^) 
X*  —  Aax^  +  6aV  —  Aa^x  +  a\  This  arrangement  would  give 
x"^  —  2ax  +  a^  for  the  quotient,  which  is  essentially  the  same  aa 
before. 


A  LITTLE   ABOUT   FACTORING.  xli 

Ex.  2.  Divide  x\f  -V  x'  -\-  if  by  xij  -^  x^  ^  if. 

Quot.,  x^  —  xy  +  f. 
Ex.  3.  Divide  a^  +  %ah  +  V  by  a  +  Z*. 
Ex.  4.  Divide  a^  -  2ah  -\' h'  hy  a  -  h. 
Ex.  5.  Divide  4:ax  +  4t"  +  a'  by  2x  +  «. 
Ex.  6.  Divide  a'b'  +  h'  +  a^b'  +  a'  by  «'  -  «J  +  ^'. 

Ex.  7.  Divide  lOac  +  3c'  +  Sa'  +  4Z»'  +  Sab  +  8Z>c  by 
25  +  rt  +  3c. 

Ex.  8.  Divide  a;'  —  y"  by  .t  —  y. 
Operation. 

x^  —  x*y 

x*y  —  If" 


x^y'^  —  y^ 


Ex.  9.  Divide  cc'  —  y'  by  a;  +  y. 

Ex.  10.  Divide  4^*  -  b'  by  2ft  -  4. 

Elx.  11.  Divide  20ft.^''  +  4.a'  -  ^x'  -  25a'x'  by  2a'  +  2x' 
—  6ax. 

Ex.  12.  Show  that  (a'  -  Z/')  -^  {a'  +  «&  +  Z»=)  =  «  -  J. 


SECTION  VII. 
A  little  about  Factoring. 

22>  Tivo  or  more  numbers  tvMch,  being  multiplied  to- 
gether,  2^'f'oduce  a  given  number,  are  called  its  factors. 
Thus  3  and  4  are  the  factors  of  12,  because  3  X  4  —  12. 


xlii  INTRODUCTION. 

So  a  -\-  h  and  a  —  b  are  the  factors  of  a'  —  b^,  because 
(a  +  b)  X  (a  —  d)  =  a''  —  b\     Try  it,  and  see. 

Ex.  1.  What  is  the  product  of  a  and  b  ?  AVhat  are  the 
factors  of  ab  ? 

Ex.  2.  What  is  the  product  of  3,  x,  and  ?/  ?  What  are  the 
factors  of  dxy  ? 

Ex.  3.  What  does  «'  mean  ?  What  are  the  factors  of  a^  ? 
What  of  a'b'  ?  of  5a'b''  ? 

SuG. — Such  an  expression  as  5a^b'^  may  be  resolved  in  a  great 
variety  of  ways  :  thus  5,  a,  a,  a,  h,  and  b  are  its  factors ;  also,  5,  a^, 
and  b^ ;  also,  5,  «^  «,  and  ¥ ;  also,  5a,  a^  b,  and  b,  etc. 

Ex.  4.  Wliat  is  the  product  of  a  and  x  -^  y?  AVliat  are 
the  factors  of  ax  +  ay? 

Ex.  5.  What  is  the  product  of  Sa  and  a  —  b?  What  are 
the  factors  of  3a'  -  dab  ?  of  2a  -  2ab  ? 


THE   SQUARE   OF  THE   SUM   OF  TWO   NUMBERS. 

Ex.  1.  What  is  the  product  of  a  -h  b  and  a  +  b?    What 
are  the  factors  of  a'  +  2ab  +  b"^? 

Ex.  2.  What  is  the  product  of  x  +  y  and  x  +  y?    What 
are  the  factors  of  x'^  +  2xy  +  y'? 

Ex.  3.  What  is  the  product  of  1  +  a:  and  1  +  x?     What 
are  the  factors  of  1  +  2:c  +  a;*  ? 


23.  We  see  from  the  last  examples  that  tJie  square  of  the 
Slim  of  two  mimbers  equals  the  square  of  one  of  the7n,+  twice 
the  product  of  the  two,  +  the  square  of  the  second.  Thus 
(a  +  b)  X  (a  -\-  b)  is  the  square  of  the  sum  of  the  two 
numbers  a  and  Z»,  and  is  equal  to  a"^  +  2ab  +  b'. 

This  principle,  and  those  in  24  and  25,  are  of  great  im- 
portance in  factoring. 


A  LITTLE   ABOUT  FACTORING.  xliii 

THE   SQUARE   OF   THE   DIFFERENCE   OF  TWO   NUMBERS. 

Ex.  1.  What  is  the  product  oi  x  —  y  aud  x  —  y'i  What 
are  the  factors  of  x^  —  Ixy  +  ?/*  ? 

Ex.  2.  AVhat  is  the  product  of  m  —  n  and  m  —  n'i  What 
are  the  factors  of  i)f  —  2mn  +  if  ? 

Ex.  3.  What  is  the  product  of  1  —  a;  and  1  —  a;  ?  What 
are  the  factors  of  1  —  2x  +  x^  ? 

Ex.  4.  What  is  the  product  of  2  -  re  and  2  —  a;  ?  What 
are  the  factors  of  4  —  4:X  •{■  x^'^ 

24z.  From  these  examples,  we  see  that  the  square  of  the 
difference  of  two  numbers  is  equal  to  the  square  of  one  of 
them,—  tio ice  the2)roduct  of  the  two,+  the  square  of  the 
other.     Thus  (x  -  y)  X  {x  —  y)  =  x*  —  '^xy  +  ^'. 

THE   PRODUCT   OF   THE    SUM  AND   DIFFERENCE   OF  TWO 
NUMBERS. 

Ex.  1.  What  is  the  product  oi  x  -\- y  and  a;  —  y  ?  What 
are  the  factors  of  x^  —  y^'- 

Ex.  2.  What  is  the  product  of  a  +  I)  and  a  -  b?  What 
are  the  factors  of  <^  —  V'^. 

Ex.  3.  What  is  the  product  of  1  +  a:  and  1  —  a;  ?  What 
are  the  factors  of  1  —  ic^  ? 

Ex.  4.  What  is  the  product  of  2  +  ic  and  2  -  a;  ?  What 
are  the  factors  of  4  —  x^  ? 

26.  We  see,  from  these  examples,  that  the  product  of 
the  sum  and  difference  of  two  numbers  is  equal  to  the  diff'er- 
ence  of  their  squares.    Thus  (x  -\-  y)  X  (x  —  y)  =  x^  —  y\ 


Ex.  1.  What  are  the  factors  of  2a  -  2&.? 
Ex.  2.  What  are  the  factors  of  3fl'  -  ^a^'x  ? 
Ex.  3.  What  are  the  factors  of  &  +  led  +  (/'  ? 


Xliv  INTRODUCTION. 

Ex.  4.  What  are  the  factors  of  a^  —  2am  +  vv"  ? 

Ex,  5.  What  are  the  factors  of  a^  —  c^  ? 

Ex.  6.  What  are  the  factors  of  of  —  2x  +  1  ? 

Ex.  7.  What  are  the  factors  of  9  -  a;'  ? 

Ex.  8.  What  are  the  factors  of  «'  +  2a  +  1  ? 


SECTION  VIII, 

How  Operations  in  Fractions  are  performed  in  the 
Literal  Notation. 

2B.  For  the  various  operations  in  fractions  in  the 
literal  notation,  the  ordinary  rules  of  arithmetic  for  the 
corresponding  cases  apply,  only,  that  the  fundamental 
operations  of  addition,  subtraction,  multiplication,  and 
division  are  performed  by  the  preceding  rules. 


TO   REDUCE   FRACTIONS  TO  THEIR   LOWEST  TERMS. 

27*  What  is  the  rule  for  this  operatmi  in  arithmetic  9 

Ex.  1.  Reduce  — -„^  to  its  lowest  terms. 

loam  X 

Result,  -X — . 
omx 

105^'v' 
Ex.  2.  Eeduce    ^  ^    '{   to  its  lowest  terms. 
looy 

Ex.  3.  Reduce  -^--^ — .    ,   ..o  to  its  lowest  terms. 
Wbx  +  4a'^' 

SuG, — Divide  numerator  and  denominator  by  4a^5. 

Ex.  4.  Reduce ^-r-^ -^  to  its  lowest  terms. 

Ix^y 

Ex.5.  Reduce ^^^r:r^^-^ to     its    lowest 

terms. 


OPERATIONS  IN  FRACTIONS  IN  THE  LITERAL  NOTATION,      xlv 


^2 X^ 

Ex.  6.  Reduce  -5 — -^ 5  to  its  lowest  terms. 

a   +  2ax  +  X 

SuG. — Try  a  +  x  and  see  if  it  will  not  divide  both  terms  of  the 
fraction. 

Ex.  7.  Eeduce  -^ ,-^  to  its  lowest  terms. 

a  —  b 

Result 


a  —  h 


Ex.  8.  Eeduce  — ^ c i  to  its  lowest  terms. 

m  —  Z}}i)i  4-  n 

SuG. — Will  any  monomial  divide  tlie  terms  of  tlie  fraction  ? 


IMPROPER  FRACTIONS   REDUCED   TO  WHOLE   OR  MIXED  . 
NUMBERS.  — /^ 

28.   What  is  the  rule  for  this  02)eratio}i  in  arithmetic? 

Ex.  1.  Eeduce ^ to  an  integral  form. 

X  —  a  ^ 

Result,  X  —  a. 
SuG.— Divide  tlie  numerator  by  the  denominator. 

T-i      ft    T»   1         10«'  —  ^^ax  —  Sx^  ,  .    ,         ,  „ 

Ex.  2.  Reduce 7 ^^ to  an  integral  form. 

11a  —  dx 

-^      ^    ^  ,        12c'  +  Sac^x"  -  3acx  -  2ftV    , 

Ex.  3.  Reduce r-^ to  an    m- 

4c  —  ax 

tegral  form. 

Ex,  4.  Reduce — to  an  integral  form. 

oa  +  X 

T.     .    -r^  -,         12c'  +  Sac'^x^  —  3aox  -  2a^x'    ^ 

Ex.  5.  Reduce  ^ — --: — s to   an   m- 

00  4-  2ax 

fegral  form. 
Ex.  6.  Reduce  ^-^ to  an  integral  or  mixed  form. 


Xlvi  INTRODUCTION. 

Operation.     ^'~ — -.    The  term  ay  can  be  divided  by  y,  giving 

a  +  - 

y 

a.    But  we  can  only  indicate  llie  division  of  b  by  y,  by  writing?  it  in 
the  form  -,  and  as  the  sign  of  both  b  and  2/  is  + ,  the  quotient  -  is  + , 

y  y 

and  is  to  be  added  to  a. 

'jlOx^ XOx  4-  4 

Ex.  6.  Reduce  — ^ to  an  intes^ral  or  mixed 

ox  ^ 

4 
form.  Result,  40:"  —  2  +  — . 

ox 

2a^x  —  x^ 
Ex.  7.  Reduce  — ^ to  an  intesfral  or  mixed  form. 

x^ 

Result,  2ax . 

a 

Query. — Why  has  —  the  —  sign  ? 

Q^   4-  ^'  ^* 

Ex.  8.  Reduce '- to  an  intesjral  or  mixed  form. 

X* 

Result,  «'  —  ax  ■}-  x^  — 


a  +  X 


Ex.  9.  Reduce  — to  an  integral  or  mixed  form. 

X  ° 

-,      ah  —  y^ 
Also, . 


MIXED   NUMBERS   REDUCED    TO   IMPROPER   FRACTION'S. 

29'   What  is  the  rule  for  this  operation  in  arithmetic  9 

Ex.  1.  Reduce  a to  the  form  of  a  fraction. 

a 

Operation.— The  integral  part  is  a.     This  multiplied  by  a  makes 

a^    From  this  b  is  to  be  subtracted,  because  of  the  —  sign.    Thus 

^  a'^  -b 

we  have  for  the  result -. 

a 


OPERATIONS  IN  FRACTIONS  IN  THE  LITERAL  NOTATION,    xlvii 

Ex.  2.  Eeduce  2a:  +  —  to  the  form  of  a  fraction.  Also, 
a  —  X . 

X 

ci^  4-  '^cix  +  x^ 
Ex.  3.  Eeduce  a  +  x ~ to  the  form  of  a 

a  —  X 

fraction. 

Sug's. — The  integral  part  a  -^^  x,  multiplied  by  the  denominator 
a  —  X,  gives  a^  —  x^.  Now  notice  that  the  numerator 
a'  +  2aa;  +  a-'^  is  to  be  subtracted  from  this,  as  the  sign  before  the 
fraction  is  — ,     If  we  subtract  a"^  +  2ax  +  x^  from  a^  —  x^,  we  have 

—  'lax  —  2^1       Hence  the  result  is  — - — — . 

a  —  x 

We  may  also  write  this  thus  : 

a""  +  2ax  +  x^       a^  -  x^  -  (a^  +  2ax  +  x^)       ^^ 

a  +  X = ^ ^.     Now   the 

a  —  X  a  —  X 

quantity  in  the  parenthesis  is  to  be   subtracted  from  the  a^  —  x^. 

Hence,  changing  the  signs  of  the  terms  in  the  parenthesis,   and 

q}  2;^ Q^  2ax  ic'^ 

dropping  the  marks,  we  have .      Adding 

a  —  X 

<>ax ^x"^ 

similar  terms,  this  becomes  — "^^ . 


Ex.  4.  Eeduce  1 5-— — ^i to  the  form  of  a  frac- 

a   +  0 

tion.  EesuUf  -j 


—  to  the  i 
a  +  X 


Ex.  5.  Eeduce  a  —  x '- to  the  form  of  a 


fraction.  Result, . 


4^2  __  g 

Ex.  6.  Eeduce  Zx  —  — ^ to  the  form  of  a  fraction. 

ox 


Result,  —- -. 

bx 


xlviii  INTRODUCTION. 

TO  REDUCE  FRACTIONS  TO   EQUIVALENT    FRACTIONS    HA7' 
ING   A   COMMON   DENOMINATOR. 

SO.   Give  the  rules  of  arithmetic  for  tins  process. 

"VYe  shall  use  only  the  method  of  multiplying  both  terms 
of  each  fraction  by  the  denominators  of  all  the  other 
fractions. 

Ex.  1.  Reduce  -,  -,  -   to  equivalent  fractions  having  a 

,  .     ,  7^      ,,    ciyz  bxz   cxy 

common  denommator.  Results,  -^^,  — ,  — -. 

xyz  xyz  xyz 

Ex.  2.  Reduce  7->  —  ?  t  to  equivalent  fractions  having  a 

common  denominator.    Also,  — ; — , ,  j^. 

X  i-  y  X  —  y  3 

Mesults  of  the  la^f  ^^^  ~  ^"^^  2^^  ^^-^  ^^'  -  ^V" 
Uesults  of  the  last,  ^^^^r^^'  Z^T^^^  ^^^'W^ 


TO   ADD     FRACTIONS. 

SI.  Repeat  the  rules  of  arithmetic  for  this  purpose, 

Ex.  1.  Add  -,  -,  and    -.  Sum,  -^r^. 

Zoo  ou 

Ex.  2.  Add  — -,   and  — - — .  Siwi,  — - — . 

o  2  o 

Ex.  3.  Add  — - — ,  and ^r — .  Sum, 

Ex.  4.  Add 7,  and 7.  Swn, 

a  —  V  a  -h  b 

Ex.  5.  Add  Y^„  and   f^^-  ^um,  ^  _  ^, 


'     10    • 

2a'  +  2b' 

a'  -b'  ' 

2  +  2:c* 

OPEKATIONS  IN  FEACTIONS  IN  THE  LITERAL  NOTATION,    xlix 


TO   SUBTRACT  FRACTIONS. 

32*   Wliat  is  the  rule  given  in  aritJimetic  for  this  pur- 
pose  f 

Ex.  1.  From  —  take  -.,  Rem.,  — . 

3  4  '12 

Ex.  2.  From  I  take  |.  Rem.,  — ^. 

Ex.  3.  From  — - —  take  — - — .  Rem.,  — — — . 

6  9  15 


Ex.  4.  From take  .  Rem.,  ^ ^. 

x-y            x+y  ^   - y 

"[  —   x^              1   +  x^  —  4a:' 

Ex.  5.  From  -— — ^   take =.       Rem., V- 

1    +    X                     1    —    X  •    1  —  £C* 


MULTIPLICATION  OF  FRACTIONS. 

33.  How  is  a  fraction  multi2)lied  by  a  ivhole  nitmber? 
Ex.  1.  Multiply  ll  by  ha\  Prod.,  ^'. 

Ex.  2.  Multiply  If^^  by  a  H-  I>.  Prod.,  |^. 

Ex.  3.  Multiply  1-^  by  1  -  a;.  Prod.,  \^. 

-L  —  X  1  -]-  X 

Ex.  4.  Multiply  ^  by  3a.  Prod.,  ^. 

Ei.  5.  Multiply  3a;  by   ^.  Pro(?.,  ^. 


1  I^TBODUCTION. 

Ex.6.Multiplyg^|5^|by«  +  J. 

„     ,      a'  —  ^ah-^V. 
Prod.,   -J . 

Ex.  7.  Multiply  10a'  by  j^.  Prod.,  — . 

Ex.  8.  Multiply  I  by  6 ;  by  8  ;  by  10. 

/V 

Ex.  9.  Multiply  ^~  by  3  ;  — ^  by  5. 

O  0 


34:.  Give  the  rule  for   muUij^lying    one    fraction  ly 
another. 

Ex.  1.  Multiply  J  by  ^.  P''^-'W 

Ex.2.  Multiply^  by-.  ^'•'"'•'57/ 

Ex.  3.  Multiply  i^  by  1±^.  ProA.,  -^^^., 
Ex.  4.  Multiply  ?^  byl-5?.  Prod,  1^. 

Ex.  5.  Multiply  -y-^by  ;^— ^.  Prod.,        3^       • 


DIVISION   OF   FKACTIONS. 

5*5^.  ^o?(;  15  «  fraction  divided  ly  a  whole  number  f 
Ex.  1.  Divide   —  by  ^x".  Quot,  --^. 

ox  i-VX 

Ex.  2.  Divide  -^-3  by  2a^b,  Quot.,  j^. 


HOW  PBOBLEMS  ARE  SOLVED  IN  ALGEBRA.      ]i 


Ex.  3.  Divide  ^^—-^  hj  a-h  Quot,  ^^^. 

Ex.  4.  Divide  -^-^  \^ix-y.  Quot,  ^^—--,, 


36.  Hoiu  is  any  quantify  divided  by  a  fraction  ? 
Ans.  By  rnuUiplyiny  it  by  the  divisor  inverted. 

Ex.  1.  Divide  6  by—.  Quot.,  — . 

Ex.  2.  Divide  ^  by  ?  Quot.,  f?. 

0  X  lo 

Ex.  3.  Divide  — 5—  by .  Quot.,     .,.    . 

Ex.  4.  Divide  ^  by . 

c  +  d    -^    X  +  y 

ax  -\-  ay  —  X  —  y 


Quot.. 


cm  +  dm  —  en  —  dn 


Ex.  5.  Divide ^ — —  by       ,      . 

ab  *'       be 


„     ^    ex  —  cy 
Quot.,  ^ 


a 


Ex.  6.  Divide ^  by  -^ 7^.  C^fO^.,  — 5 . 

a  +  b    "^  a  —  b  ox 


SECTION  IX. 

How  Problems  are  Solved  in  Algebra. 

Ex.  1.  John  is  3  times  as  old  as  James,  and  the  sum  of 
their  ages  is  32 ;  how  old  is  each  ? 


lii  INTRODUCTION. 

Solution. — This  example  is  a  very  simple  one,  and  can  easily  be 
solved  mentally.  Thus,  we  see  that  since  John's  age  is  3  times 
James's,  both  their  ages  together  make  4  times  James's  age,  and  this 
is  33  years.  Now  4  times  James's  age  =  32  years.  Hence,  James's 
age  is  i  of  32,  or  8  years ;  and  John's  age  being  3  times  James's,  is 
3  X  8,  or  24  years. 

To  solve  this  by  Algebra,  we  proceed  as  follows  :  Let  x  represent 
James's  age  ;  then,  since  John  is  3  times  as  old,  Zx  will  represent  his 
age;  and  the  sum  of  their  ages  will  be  3^'  +  x.  Now  the  statement 
is  that  the  sum  of  their  ages  is  32;  hence  ^x  +  x  =z  32.  Or,  what  is 
the  same  thing,  Ax  =  32.  If  Ax  =  32,  «  =  i  of  32,  or  a;  =  8.  But 
X  stood  for  James's  age,  hence,  James's  age  is  §;  and  John's  being  3 
times  as  much,  is  3  x  8,  or  24. 

37'  The  ex2)ression  ^x  +  x  =  ^2  is  what  is  called  an 
Equatioii ;  and  it  is  hythe  use  of  equations  that  loe  solve 
pivblems  in  Algebra. 

Ex.  2.  A  merchant  said  that  he  had  72  yards  of  a  cer- 
tain kind  of  cloth,  in  three  rolls.  In  the  first  roll,  there 
were  a  certain  number  of  yards ;  in  the  second,  3  times  as 
many  as  in  the  first ;  and  in  the  third,  twice  as  many  as  in 
the  first.     How  many  yards  were  there  in  the  first  roll  ? 

Sug's.— The  equafion  is  a;  +  3a;  +  2^  =  72. 

Now,  X  +  dx  +  2x  is  Qx,  hence  6x  =  72. 

And  if  Qx  =  72,  x,  or  Ix,  is  i  of  72,  x  =  12. 

Queries. — What  does  the  x  stand  for  ?  Answer.  The  number  of 
yards  in  the  first  roll.  In  this  problem  which  is  the  most, a;  +  3.»  +  2a;, 
or  72  ?  To  start  with,  do  you  know  how  much  a;  is  ?  Then  is  it  a 
known,  or  an  unknown  quantity,  at  the  outset  ? 


58.— The  number  which  we  desire  to  find  as  the  answer 
of  a  problem  is  represented  in  the  beginning  of  the  solution 
by  one  of  the  latter  letters  of  the  alphabet,  usually  x,  if 
there  is  need  of  but  one  letter,  and  is  called  the  JJnUnown 
Quantity. 

Ex.  3.  A  boy  on  being  asked  how  old  he  was,  replied, 
"  if  you  add  to  my  age  3  times  my  age,  and  5  times  my  age, 


HOW   PKOBLEMS   ARE   SOLVED   IN  ALGEBRA.  liii 

and  subtract  twice  my  age,  the  result  will  be  49  years.   How 
old  was  he  ? 

Sug's.— The  equation  is  x  +  Sx  +  5x  —  2x  =  49. 

Hence,  since  x  +  dx  +  5x  —  2x  is  7x,  7x  =  49. 

If  7«  =  49,  «  =  I  of  49,  or  x  =    7. 

Ex.  4.  There  are  three  times  as  many  girls  as  boys  in  a 
party  of  60  children.  How  many  boys  are  there  ?  How 
many  girls  ? 

Ex.  5.  In  a  barrel  of  sugar  weighing  200  lbs,  there  aro 
three  varieties,  A,  B,  and  C,  mixed.  There  is  7  times  as 
much  of  B  as  of  A,  and  twice  as  much  of  0  as  of  A.  How 
much  of  A  is  there?  How  much  of  each  of  the  other 
kinds  ?     A71S.,  of  A,  20  lbs. ;  of  B,  140  lbs. ;  of  C,  40  lbs. 

Ex.  6.  There  were  4  kinds  of  liquor  put  into  a  cask,  2 
times  as  much  of  the  second  as  of  the  first,  2  times  as  much 
of  the  third  as  of  the  second,  and  2  times  as  much  of  the 
fourth  as  of  the  third.  The  cask  sprang  a  leak,  and  three 
times  as  much  leaked  out  as  was  put  in  of  the  first  kind, 
when  it  was  found  that  there  were  36  gallons  remaining. 
How  much  was  there  put  in  of  each  kind  ? 

Sug's, — The  equation  is  x  +  2x  +  4^  +  8x  —  dx  =  36. 


Ex.  7.  In  a  pasture  there  are  a  certain  number  of  cows 
and  23  sheep,  in  all  34  animals.  How  many  cows  were 
there  ? 

Solution.  As  it  is  tlie  number  of  cows  we  seek,  let  x  represent 
the  number  of  cows.  Then  x  +  23  is  the  number  of  animals  in  the 
pasture,  and  the  equation  is 

a;  +  23  =  34. 
Now  the  X  +  23  means  just  the  same  thing  as  the  34,  that  is 
05  +  23  3=  34.  So,  if  w^e  subtract  23  from  each,  there  will  be  just  as 
much  left  of  one  as  of  the  other.  Subtracting  23  from  x  +  23,  there 
remains  x,  and  subtracting  23  from  34,  there  remains  11.  Hence 
X  =  11.  Now  as  x  represented  the  number  of  cows,  we  know  that 
there  were  11  cows. 


liv  INTRODUCTION. 

39'  The  part  of  an  equation  on  the  left  of  the  sign  =  is 
called  the  First  Meuiher,  and  that  on  the  right,  the 
Second  Member. 

Note. — The  pupil  must  not  think  that  because  these  examples 
are  so  simple  that  he  can  "  do  them  in  his  head"  without  any  al- 
gebra, and  may  be  with  less  work,  that  therefore  algebra  is  a  very 
clumsy  method  of  solving  examples.  He  will  find  by  and  by,  that 
though  the  equation  does  not  really  help  any  in  the  solution  of  such 
simple  questions,  it  will  solve  a  great  many  very  difficult  ones  about 
which  he  might  puzzle  his  brains  a  great  while  to  no  purpose,  if 
algebra  did  not  come  to  his  aid.  Stick  to  it,  then,  and  learn  how  to 
use  this  new  instrument,  the  Equation,  and  you  will  by  and  by  find 
it  wonderfully  useful.     It  is  a  grand  patent  for  solving  problems. 

Ex.  8.  In  a  certain  pasture  there  are  three  times  as  many 
horses  as  cows,  and  20  sheep.  In  all  there  are  100  ani- 
mals.   How  many  cows  are  there  ?     How  many  horses  ? 

Sug's.— The  equation  is  a;  +  3a;  +  20  =  100. 

Subtracting  20  from  each  member,  a;  +  3a;  =    80. 

Uniting  the  terms  of  the  first  member,  4a;  =    80. 

Dividing  each  member  by  4,  x=    20. 

Hence  there  were  20  cows ;  and,  as  there  were  tliree  times  as  many 
horses  as  cows,  there  were  CO  horses. 

Ex.  9.  In  a  basket  of  60  apples  there  are  4  times  as  many 
red  apples  as  yellow,  and  10  green  apples.  How  many 
yellow  apples  are  there?     How  many  red  ? 

Ex.  10.  John  and  James  together  have  75  cents.  James 
has  25  cents  less  than  John.     How  many  cents  has  John  ? 

Sug's. — Let  x  represent  the  number  of  cents  which  John  has. 
Then,  as  James  has  25  cents  less,  a;  —  25  will  represent  what  he  has. 
But  both  together  have  75  cents.     Hence  the  equation  is 

a;  +  a;  —  25  =  75. 

Now,  if  we  drop  the  —  25  from  the  first  member,  we  make  this 
member  25  greater  than  it  now  is,  *.  «.,  x  +  x  is  25  greater  than 


HOW  PROBLEMS  ARE  SOLVED  IN  ALGEBRA.      Iv 

X  +  x—25.    Therefore,  if  we  add  25  to  the  Becond  member,  makmg 

it  100,  the  members  will  still  be  equal  * 

This  gives  x  -\-  x  =  100, 

or,  2i;=100. 

Hence  x  =    50,  the  number  of  cents  which 

John  has. 

Ex.  11.  A  merchant  has  90  yards  of  cloth  in  two  pieces. 
The  longer  piece  lacks  ten  yards  of  containing  3  times  as 
mnch  as  the  shorter.     How  much  in  each  piece  ? 

SuG. — The  equation  is  a;  +  ox  —  10  =  90. 

Ex.  12.  Divide  the  number  50  into  two  parts  so  that  one 
part  shall  lack  10  of  being  5  times  the  other. 

Sug's. — The  parts  are  represented  by  x^  and  5x  —  10.  They  are 
10  and  40. 

Ex.  13.  Divide  the  number  50  into  3  parts,  such  that 
the  second  shall  be  5  more,  and  the  third  15  less  than  the 
first. 

Sug's.— The  equation  is  a;  +  aj  +  5  +  a;  —  15  =  50.  The  parts  are 
20,  25,  and  5. 

Ex.  14.  There  are  52  animals  in  a  field.  Twice  the  num- 
ber of  cows  +  11  is  the  number  of  sheep,  and  3  times  the 
number  of  cows  —  13  is  the  number  of  horses.  How  many 
of  each  kind  ?         Aus.,  9  cows,  29  sheep,  and  14  horses. 


Ex.  15.  A  man  said  of  his  age, 
"  If  to  my  age  there  added  be 
One  half,t  one  third,  and  three  times  three. 
Six  score  and  ten  the  sum  will  be. 
What  is  my  age  ?     Pray  show  it  me." 

Sug's.— The  equation  isa:  +  ^  +  5  +9  =  130. 

*  The  word  "  Transposition"  is  pui-posely  omitted  from  this  introduction.  Nor 
is  the  idea  designed  to  be  presented.  It  will  be  better  for  the  pupil  to  "  think 
out"  the  process,  as  above. 

+  Meaning  "  one  half  my  aj/e,"  "  one  third  my  age.'^ 


Ivi  INTRODUCTION. 

Subtracting  9  from  each  member- 

..|. 1  =  121. 

Now,  we  can  get  rid  of  the  fractions  in  the  first  member  by  mul- 
tiplying it  by  6,  the  product  of  both  the  denominators.  Thus,  6 
times  the  first  member  is  Qx  +  Sx  +  2x.  Then,  if  we  also  multiply 
the  second  member  by  6, the  products  will  be  equal.  For  if  two 
quantities  are  equal,*  6  times  one  of  them  is  equal  to  6  times  the 
other.  Hence  we  have  6x  +  dx  +  2x  =  726. 
Uniting  terms,  lla;  =  726. 

Dividing  by  11,  a:  =    66. 

Ex.  16.  Mary  gave  half  her  books  to  Jane,  and  one  third 
of  them  to  Helen,  when  she  had  but  ^  left.  How  many 
had  she  at  first  ? 

Sug's. — Let  X  represent  the  number  of  books  Mary  had  at  first. 

Then  she  gave  Jane  -,  and  Helen  -  books.     And  what  she  gave 

the  otlier  girls,  added  to  what  she  had  left,makes  all  she  had  in  the 
first  place.     Hence  the  equation  is 

XX. 

2  +  3  +  ^  =  "- 

Multiplying  each  member  by  6,  dx  +  2x  +  12  =  Gx. 

Subtracting  5a;  from  each  member,  13  —    x. 

That  is,  Mary  had  12  books  at  first. 

Ex.  17.  A  boy  lost  25  cents  of  some  money  which  his 
uncle  gave  him,  and  gave  half  he  had  left  to  his  brother. 
He  then  earned  50  cents,  when  he  had  just  as  much  as  his 
uncle  gave  him.     How  much  did  his  uncle  give  him  ? 

Sug's. — Let  x=\  the  number  of  cents  his  uncle  gave  him.  Then 
he  had  a;  —  25  cents  after  losing  25  cents.    After  giving  away  half 

of  this,  he  had  the  other  half,  or  — - —  cents,  left.    He  then  earned 

2 

50  cents,  and  the  amount  he  had  was  equal  to  what  his  uncle  gave 
him, 

*  In  this  case  the  two  quantities  are  the  two  members. 

t  The  Bi^n  of  equality  used  in  this  way  means  the  same  as  the  word  "  reprc- 
eent." 


HOW   PROBIJSMS   ARE   SOLVED   IN  ALGEBRA.  Ivii 

Hence  the  equation  is  — ^— ^  +  50  =    x. 

Multiplying  each  member  by  2,  a:  —  25  +  100  =  2x. 

Uniting,  —  25  +  100  makes  75,  and  x  +  75  =  2x. 

Subtracting  x  from  each  member,  75  =    x. 

Ex.  18.  A  boy  being  asked  how  many  marbles  he  had, 
said,  "  If  I  had  five  more  than  I  have,  half  the  number 
subtracted  from  30  would  leave  twice  as  many  as  I  now 
have."     How  many  marbles  had  he  ? 

Sxtg's. — Letting  x  represent  the  number  of  marbles  the  boy  had, 
the  equation  is 

Now  there  is  a  little  peculiarity  about  this  equation,  which  the 
pupil  must  be  careful  to  notice  whenever  it  occurs,  or  he  will  make 
a  great  many  mistakes.  It  is  this  :  When  we  multiply  both  mem- 
bers by  2,  to  get  rid  of  the  fraction,  we  must  write  60  —  x  —  5  =  4x 
The  mistake  would  be  to  write  60  —  :r  +  5  =  4a:.     The  explanation 

is,  that  the  —  sign  before  — - —  shows  tliat  it  is  to  be  subtracted  from 

30,  hence  the  signs  of  the  terms  composing  it,  viz.,  x  and  5,  must  be 
changed,  according  to  tbe  rule  for  subtraction.  But  the  pupil  may 
think  that  we  do  not  change  the  sign  of  the  x.    If  he  does  he  mis- 

2/4-5 

takes.    The  —  sign  before  the  fraction  — ^ —  does  not  belong  to  the 

X,  but  to  the  fraction  as  a  whole.      The  sign    of  x  in  the  fraction 

— is  +,  since   when  no  sign  is  expressed   +   is  understood. 

2 
"What  then  becomes  of  the  —  sign  before  the  fraction,  if  it  is  not  the 
same  as  the  sign  of  a;  in  the  equation  60—  ic  —  5  =  4^?  It  has  been 
dropped,  since  the  thing  signified  by  it  has  been  performed,  and 
the  —  sign  before  the  x  is  the  sign  of  that  term  in  the  original  equa- 
tion, changed.  "    The  boy  had  11  marbles. 

Ex.  19.   What    is    the    value    of    x    in    the   equation 

Sug's. — Multiplying  each  member  by  3,  we  have  Ox]—  2  +  22;  =  Gl. 
Hence  »  =  6. 


Iviii  INTBQDUCTION. 

Ex.  20.      Find     the    value     of    x    in    the     equation 

3a:  -  5       ,  ,        2a;  -  4 
X  +  — ~ —  =  12 ^ — .  X  =  ^. 

Ex.  21.    What    is    the    value    of   x    in    the   equation 

X  —  1      X  -\-  4:      ^^       X  +  3  ^ 

Ex.    22.   Show   thiit   in  ^^-^^  = 


Ans., 

X 

=  40A. 

23- 

X 

4  + 

X 

5 

4 

~i 

Ex.  23.  Two  boys  were  to  divide  32  marbles  between 
them  so  that  ^  of  what  one  had  should  be  5  less  than  what 
the  other  had.     How  many  was  each  to  have  ? 

Sug's. — Letting  x  ■=  what  one  had, 
then  33  —  a:  =  what  the  other  had. 

X 

The  equation  is  -  +  5  =  32  —  ar, 


2 
Z'l-x 


or 


+  5  =  a:. 


2 
Query. — Whj''  will  either  equation  answer  the  purpose  ? 

Ex.  24.  AVhat  number  is  that  to  which  if  7  be  added, 
half  the  sum  will  exceed  \  of  the  remainder  of  the  number 
after  3  has  been  subtracted,  by  8  ? 

^.  X  + 1     x  —  Z     ^ 

Equation,  — z —  =  o.  x—  id. 

Ex.  25.  The  sum  of  two  numbers  is  sixteen,  and  tlie 
less  number  divided  by  three  is  equal  to  the  greater  divided 
by  five.     What  are  the  numbers  ? 

SuG.— Let  x and  16  —  x  represent  the  numbers. 

Ex.  2G.  Divide  twenty-two  dollars  between  A  and  B,  so 
that  if  one  dollar  be  taken  from  three-fourths  of  B's  share, 
and  three  dollars  be  added  to  one-half  of  A's  money,  the 
sums  shall  be  equal.     How  many  dollars  will  each  have  ? 


HOW  PROBLEMS  ARE  SOLVED  IN  ALGEBRA.      lix 

Ex.  27.  The  sum  of  two  numbers  is  thirty-three.  If 
one-sixth  of  the  greater  be  subtracted  from  two-thirds  of 
the  less  number,  the  remainder  will  be  seven.  What  are 
the  numbers  ? 

Ex.  28.  The  sum  of  A's  and  B's  money  is  thirty-six 
dollars.  If  five-eighths  of  B's,  less  two  dollars,  be  taken 
from  three-fourths  of  A's,  the  difference  will  be  seven  dol- 
lars.    How  many  dollars  has  each  ? 

Ex.  29.  The  difference  between  two  numbers  is  twenty -five: 
and  if  twice  the  less  be  taken  from  three  times  the  greater, 
the  remainder  will  be  eighty.     What  are  the  numbers  ? 

Ex.  30.  A  and  Bgain  money  in  trade,  but  A  receives  ten 
dollars  less  than  B.  If  A's  share  be  subtracted  from  twice 
B's,  the  remainder  will  be  fifty-seven  dollars.  How  much 
money  did  each  receive  ? 

Ex.  31.  One  number  is  four  less  than  another,  and  if 
twice  the  less  be  subtracted  from  five  times  the  greater,  the 
remainder  Avill  be  thirty-eight.     What  are  the  numbers? 

Ex.  32.  Two  farms  belong  to  A  and  B.  A  has  twenty 
acres  less  than  B.  If  twice  A's  farm  be  taken  from  three 
times  B's  number  of  acres,  the  remainder  will  be  one  hun- 
dred acres.     How  many  acres  has  each  ? 

Ex.  33.  One  number  is  seven  less  than  another,  and  if 
three  times  the  less  be  taken  from  four  times  the  greater, 
the  remainder  will  be  six  times  the  difierence  between  the 
two  numbers.     What  are  the  numbers  ? 

Ex.  34.  Anna  is  four  years  younger  than  Mary.  If  twice 
Anna's  age  be  taken  from  five  times  Mary's,  the  remainder 
will  be  thirty-five  years.    What  is  the  age  of  each  ? 

Ex.  35.  One  number  is  ten  less  than  another.  If  three 
times  the  less  be  taken  from  five  times  the  greater,  the  re- 
mainder will  be  seven  times  the  difference  of  the  two  num- 
bers.    What  are  the  numbers  ? 


INTRODUCTION. 


SECTION  L 

A  Brief  Survey  of  the  Object  of  Pure  Mathematics  and 
of  the  several  Branches. 

[Note.  —Omit  this  section  till  a  General  Keview  is  taken.  ] 

1,  I* are  Mathematics  is  a  general  term  applied  to 
several  branches  of  science,  which  have  for  their  object  the 
investigatioix  of  the  properties  and  relations  of  quantity 
— comprehending  number,  and  magnitude  as  the  result  of 
extension — -and  of  form. 

2.  The  Several  branches  of  Pure  Mathema- 
tics   are  Arithmetic,  Algebra,  Calculus,  and  Geometry. 

5.  Arithmetic,  Algebra,  and  Calculus  treat  of  number, 
and  Geometry  treats  ^f  magnitude  as  the  result  of  exten- 
sion. 

4.  Quantity  is  the  amount  or  extent  of  that  which 
may  be  measured  ;  it  comprehends  number  and  magnitude. 

The  term  quantity  is  also  conventionally  apphed  to  sym- 
bols used  to  represent  quantity.  Thus  25,  m,  xi,  etc.,  are 
called  quantities,  although,  strictly  speaking,  they  are  only 
representatives  of  quantities. 

It  is  not  easy  to  give  a  philosophical  acconnt  of  the  idea  or  ideas, 
represented  by  the  word  Q'lanUiy  as  used  in  Mathematics  ;  and 
doubtless,  diflferent  persons  use  the  word  in  somewhat  diflferent  senses. 
It  is  obviously  incorrect  to  say  that  "Quantity  is  anything  which  can 


2  INTRODUCTION. 

be  measured."  Quantity  may  be  afErmed  Of  any  such  concept, 
nevertheless,  it  is  not  the  thing  itself,  but  rather  the  amount  or  extent 
of  it.  Thus,  a  load  of  wood,  or  a  piece  of  ground,  can  be  measured  ; 
but  no  one  would  think  of  the  wood  or  piece  of  ground  as  being  the 
quantity.  The  quantity  (of  wood  or  ground)  is  rather  the  amount  or 
extent  of  it. 

The  word  is  very  convenient  as  a  general  term  for  mathematical 
concepts,  when  we  wish  to  speak  of  them  without  indicating  whether 
it  is  number  or  magnitude  that  is  meant.  Thus  we  say,  "m  repre- 
sents a  certain  quantity, "  and  do  not  care  to  be  more  specific. 

As  applied  to  number,  perhaps  the  term  conveys  the  idea  of  the 
whole,  rather  than  of  that  whole  as  made  up  of  parts.  It  is,  therefore, 
scarcely  proper  to  speak  of  multiplying  by  a  quantity ;  we  should  say, 
by  a  number.  On  the  other  hand,  when  we  apply  the  term  quan- 
tity to  magnitude,  it  is  with  the  idea  that  magnitude  may  be  meas- 
ured, and  thus  expressed  in  number. 

The  distinction  between  quantity  and  number  is  marked  by  the 
questions,  "  How  much?"  and  "  How  many  ?" 

S,  dumber  is  quantity  conceived  as  made  up  of  parts, 
and  answers  to  the  question,  "  How  many  ?" 

Thus,  a  distance  is  a  quantity ;  but,  if  we  call  that  distance  5,  we 
convert  the  notion  into  number,  by  indicating  that  the  distance  under 
consideration  is  made  up  of  parts.  Now,  the  distance  may  be  just 
the  same,  whether  we  consider  it  as  a  whole,  or  think  of  it  as  5  ;  i.  e. , 
as  made  up  of  5  equal  parts.  Again,  m  may  mean  a  value,  as  of  a 
farm.  We  may  or  may  not  conceive  it  as  a  number  (as  of  dollars. ) 
If  we  think  of  it  simply  in  the  aggregate,  as  the  worth  of  a  farm,  m 
represents  quantity  ;  if  we  think  of  it  as  made  up  of  parts  (as  of  dol- 
lars) it  is  a  number. 

G.  Number  is  of  two  kinds,  Discontinuous  and 
Contimiotis. 

7.  T>lscontinuous  JViiniber  is  number  conceived 
as  made  up  of  finite  parts  ;  or  it  is  number  which  passes 
from  one  state  of  value  to  another  by  the  successive  addi- 
tions or  subtractions  of  finite  units  ;  i.  e.,  units  of  appreci- 
able magnitude. 


THE   OBJECT  OF   PUEE   I^IATHEMATICS.  3 

8,  Continuous  Wumher  is  number  which  is  con- 
ceived as  composed  of  infinitesimal  parts  ;  or  it  is  number 
which  passes  from  one  state  of  vahie  to  another  by  passing 
through  all  intermediate  values,  or  states. 

Number,  as  the  pupil  has  been  accustomed  to  consider  it  in  Arith- 
metic, and  as  he  will  contemplate  it  in  this  volume,  is  Discontinuous 
Number.  Thus  5  grows  till  it  becomes  9,  by  taking  on  additions  of 
units  of  some  conceivable  value  ;  as  when  we  consider  it  as  passing 
thus,  5,  5+1  or  6,  6+1  or  7,  7+1  or  8,  8+1  or  9.  If  the  increment 
were  any  fraction,  however  small,  the  form  of  the  conception  loould  he  the 
same. 

As  to  Continuous  Number,  this  is  not  the  place  for  a  full  considera- 
tion of  the  idea  ;  hence,  only  a  single  illustration  will  be  given. 
Time  affords  one.  We  usually  conceive  time  as  a  disco7itinuous  num- 
ber, as  when  we  think  of  it  as  made  up  of  hours,  days,  weeks,  etc. 
But  it  is  easy  to  see  that  such  is  not  the  way  in  which  time  actually 
grows.  A  period  of  one  day  does  not  grow  to  be  a  period  of  on© 
week  by  taking  on  a  whole  day  at  a  time,  or  a  whole  hour,  or  even  a 
whole  second.  It  grows  by  imperceptible  increments  (additions). 
These  inconceivably  small  parts,  by  which  time  is  actually  made  up, 
we  call  infinitesimals,  and  number,  when  conceived  as  made  up  of 
such  infinitesimals,  we  call  Continuous  Number. 

9,  Arithinetic  treats  of  Discontinuous  JV^um- 

her, — of  its  nature  and  properties,  of  the  various  methods 
of  combining  and  resolving  it,  and  of  its  appUcation  to 
practical  affairs. 

The  leading  topics  of  Arithmetic  are  : 

1.  Notation;  i.  e.,  methods  of  representing  number,  as  by  the  Ara- 
bic Characters,  1,  2,  3,  4,  etc.,  or  by  letters,  as  a,  b,  m,  n,  x,  y,  etc., 

2.  Properties  of  Numbers  or  deductions  from  the  methods  of  Nota- 
tion, 

3.  Reduction,  as  from  one  scale  to  another,  from  one  denomination 
to  another,  from  one  fractional  form  to  another,  or,-  in  short,  from  any 
one  form  of  expression  to  another  equivalent  form, 

4.  The  various  methods  of  combining  number,  as  by  addition, 
multiplication,  and  involution, 


4  INTBODUCTION. 

5.  Resolving  number,  as  by  subtraction,  division,  and  evolution, 

And  all  these  processes  as  effected  by  the  use  of  any  notation,  and 
upon  integral  or  fractional  discontinuous  numbers  of  any  kind. 

Arithmetic,  therefore,  philosophically  considered,  embraces  much 
that  is  usually  classed  as  Algebra.  Thus  all  that  usually  precedes 
Simple  Equations,  and  all  that  is  embraced  in  this  Part  I. ,  is  simply 
a  repetition  and  extension  of  the  processes  of  Arithmetic  with  a  new 
notation — the  literal.  Again,  logarithms  are  nothing  but  a  new 
scheme  of  notation,  by  means  of  which  certain  combinations  and  res- 
olutions are  more  readily  effected;  and  the  making  of  logarithms  is 
but  a  reduction  from  one  form  of  expression  to  an  equivalent  one  in 
another  notation.  In  the  ordinary,  or  decimal  notation,  a  certain 
number  is  represented  thus,  256  ;  in  the  logarithmic  notation  it  is 
2.40824. 

On  the  other  hand,  much  that  is  ordinarily  embraced  in  Arithmetic 
is  more  philosophically,  and  more  economically  transferred  to  Alge- 
bra, as  will  appear  in  the  next  article. 

10*  Algebra  treats  of  the  Uquaiion,  and  is  chiefly 
occupied  in  explaining  its  nature  and  the  methods  of 
transforming  and  reducing  it,  and  in  exhibiting  the 
manner  of  using  it  as  an  instrument  for  mathematical 
investigation. 

The  whole  province  of  the  relations  of  quantity,  continuous  or  dis- 
continuous number,  is  covered  by  Algebra,  so  far  as  the  equation  can  be 
made  the  instrument  of  investigation.  Much,  therefore,  of  what  is 
found  m  our  Arithmetics  can  be  more  expeditiously  treated  by  Alge- 
bra. Such  are  the  subjects  of  Ratio,  Proportion,  the  Progressions, 
Percentage,  Alligation,  etc.  In  fact,  the  equation  is  the  grand  instru- 
ment of  mathematical  investigation,  and  demonstrates  its  efficiency  in  every 
department  of  the  science.  To  hope  to  get  on  in  mathematics  without 
Algebra,  is  to  expect  to  walk  without  feet. 

11,  Calculus  treats  of  Continuous  Number,  and  is 
chiefly  occupied  in  ■  deducing  the  relations  of  the  infini- 
tesimal elements  of  such  number  from  given  relations 
between  finite  values  and  the  converse  process,  and  also 
in  pointing  out  the  nature  of  such  infinitesimals  and  the 
methods  of  using  them  in  mathematical  investigation. 

12,  Geometry  treats  of  magnitude  and  form  as  the 
result  of  extension  and  position. 


n 


LOGICO-MATHEMATICAL  TEBMS. 


The  priof  ipal  divisions  of  the  science  of  Geometry  are  : 

1.  The  ihicient,  Special  or  direct  Geometrj^  (the  common  Geometry 
of  our  schools,)  including  Trigonometry,  Conic  Sections,  and  all 
other  geometrical  inquiries  conducted  upon  these  methods, 

2»  The  Modern,  Indirect  or  General  Geometry  ^usually  called  An- 
alytical,)  and 

3.  Descriptive  Geometry. 


SYNOPSIS. 


Subject  of  Section. 

Definition  of  Pure  Mathematics  : 

Several  Branches  of. 
Subject    matter    of   the    several 

branches. 
Quantity  :     Two    uses     of     the 

term. 


Number  :  Illustration  of :  Kinds 
of:  Illustration  of  kinds  of 
number. 

Arithmetic  :  Topics  of. 

Algebra. 

Calculus. 

Geometry. 


SUCTION   11. 
Logico-Mathematical  Terms. 

IS.  A  JPvoposition  is  a  statement  of  something  to 
be  considered  or  done. 

Illustration. — Thus,  the  common  statement,  "  Life  is  short,"  is  a 
proposition  ;  so,  also,  we  make,  or  state  a  proposition,  when  we  say, 
"Let us  seek  earnestly  after  truth." — "The  product  of  the  divisor 
and  quotient,  plus  the  remainder,  equals  the  dividend,"  and  the  re- 
quirement, " To  reduce  a  fraction  to  its  lowest  terms,"  are  examples 
of  Arithmetical  propositions. 

14:,  Propositions  are  distinguished  as  Axioms,  TJieorems, 
Lemmas,  Corollaries,  Postulates,  and  Problems. 

lo.  A.71  Axiom  is  a  proposition  which  states  a  princi- 
ple that  is  so  simple,  elementary  and  evident  as  to  require 
no  proof. 

Illustration. — Thus,  "  A  part  of  a  thing  is  less  than  the  whole  of 
it,''   "  Equimultiples  of  equals  arc  equal,"  are  examples  of  axioms.     If 


(5  INTRODUCTION. 

any  one  does  not  admit  the  truth  of  axioms,  when  he  understands  the 
terms  used,  we  say  that  his  mind  is  not  sound,  and  that  we  cannot 
reason  with  him. 

IG,  A.  Theorem  is  a  proposition  -which  states  a  real 
or  supposed  fact,  whose  truth  or  falsity  we  are  to  deter- 
mine by  reasoning. 

Illusteation. — "  If  the  same  quantity  be  added  to  both  numerator 
and  denominator  of  a  proper  fraction,  the  value  of  the  fraction  will  be 
increased,"  is  a  theorem.  It  is  a  statement  the  truth  or  falsity  of 
which  we  are  to  determine  by  a  course  of  reasoning. 

17 •  A.  Denionstratiofi  is  the  course  of  reasoning 
by  means  of  which  the  truth  or  falsity  of  a  theorem  is 
made  to  appear.  The  term  is  also  applied  to  a  logi- 
cal statement  of  the  reasons  for  the  processes  of  a  rule. 
A  solution  tells  how  a  thing  is^  done  :  a  demonstration  tells 
why  it  is  so  done.     A  demonstration  is  often  called  2:>roo/. 

18,  A.  LeTtima  is  a  theorem  demonstrated  for  the 
purpose  of  using  it  in  the  demonstration  of  another 
theorem. 

Illustration. — Thus,  in  order  to  demonstrate  the  rule  for  finding 
the  greatest  common  divisor  of  two  or  more  numbers,  it  may  be  best 
first  to  prove  that  ' '  A  divisor  of  two  numbers  is  a  divisor  of  their  sum, 
and  also  of  their  difference."  This  theorem,  when  proved  for  such  a 
purpose,  is  called  a  Lemma. 

The  term  Lemma  is  not  much  used,  and  is  not  very  important, 
since  most  theorems,  once  proved,  become  in  turn  auxiliary  to  the 
proof  of  others,  and  hence  might  be  called  lemmas. 

10,  A  Corollary  is  a  subordinate  theorem  which  is 
suggested,  or  the  truth  of  which  is  made  evident,  in  the 
course  of  the  demonstration  of  a  more  general  theorem, 
or  which  is  a  direct  inference  from  a  proposition. 

Illustration. — Thus,  by  the  discussion  of  the  ordinary  process  of 
performing  subtraction  in  Arithmetic,  the  following  Corollary  might 
be  suggested  :  "  Subtractioa  may  also  be  performed  by  addition,  as 
we  can  readily  observe  what  number  must  be  added  to  the  subtrahend 
to  produce  the  minuend." 


LOGICO-MATHEMATICyij   TERMS.  7 

20*  A.  Postulate  is  a  proposition  which  states  that 
somethiiig  can  be  done,  and  which  is  so  evidently  true 
as  to  require  no  process  of  reasoning  to  show  that  it  i^ 
possible  to  be  done.  We  may  or  may  not  know  how  to 
perform  the  operation. 

Tt.t.ttstrattox.  — Quantities  of  the  same  kind  can  be  added  together. 

21^  A.I*rohlei7l  is  a  proposition  to  do  some  specified 
thing,  and  is  stated  with  reference  to  developing  the 
method  of  doing  it. 

IiiiitrsTRATioN. — A  problem  is  often  stated  as  an  incomplete  sen- 
tence, as,  "  To  reduce  fractions  to  a  common  denominator." 

22,  A.  Mule  is  a  formal  statement  of  the  method  of 
solving  a  general  problem,  and  is  designed  for  practical 
application  in  solving  special  examples  of  the  same  class. 
Of  coui'se  a  rule  requires  a  demonstration. 

23,  A  Solution  is  the  process  of  performing  a  prob- 
lem or  an  example.  It  should  usually  be  accompanied 
by  a  demonstration  of  the  process. 

24,  A  ScJloliuiil  is  a  remark  made  at  the  close  of 
a  discussion,  and  designed  to  call  attention  to  some 
particular  feature  or  features  of  it._ 

Illusteation. — Thus,  after  having  discussed  the  subject  of  multi- 
pncation  and  division  in  Arithmetic,  the  remark  that  ' '  Division  is 
the  converse  of  multiplication, "  is  a  scholium. 


SYNOPSIS. 


Subject  of  the  section. 
Proposition.  III. 
Varieties  of  propositions. 
Axiom.  III. 
One    who    will    not    admit 

truth  of  axioms. 
Theorem.  III. 
Demonstration.       Difference 


tweeu  a  solution  and 
stratiou. 


the 


be- 


.  demon- 


Lemma.    III.     Why  the  term  is 

unimportant. 
Corollary.   III. 
Postulate. .  III. 
Problem.     How  stated.   III. 
Rule. 
Solution. 
Scholium.   III. 


PART  L 

LITERAL   ARITHMETIC 


CHAPTER  I. 

FUNDAMENTAL  RULES. 


SECTION  L 

Notation. 

2^.  A  System  of  Notation  is  a  system  of  symbols 
by  means  of  which  quantities,  the  relations  between 
them,  and  the  operations  to  be  performed  npon  them, 
can  be  more  concisely  expressed  than  by  the  use  of 
words. 

SYMBOLS   OF    QUANTITY. 

26,  In.  Arithmetic,  as  usually  studied,  numbers  are  rep- 
resented by  the  characters,  1,  2,  3,  4,  5,  6,  7,  8,  9,  0, 
called  Arabic  figures,  or,  simply,  figures. 


*  Part  first  treats  of  the  familiar  operations  of  Addition,  Subtraction,  Multiplica- 
tion, Division,  Involution  and  Evolution,  and  the  theory  of  Fractions.  The  only  dif- 
ference between  the  processes  hero  developed  and  those  with  which  the  pupil  is 
already  familiar,  grows  out  of  the  notation.  Hence  appears  the  appositeness  of 
the  term  Literal  Arithmetic.  Hence,  also,  the  teacher  should  be  careful  that  th« 
pupil  see  the  unity  of  purpose,  aiul  the  reason  for  any  difforoiico  in  method  of  exiy 
cutiou. 


NOTATION.  9 

j^7.  In  other  departments  of  mathematics  than  Arithme- 
tic, numbers  or  quantities  are  more  frequently  representee! 
by  the  common  letters  of  the  alphabet,  a,  b,  c,  .  .  .  m, 
n,  .  .  ,  X,  y,  z.  These  letters  may,  however,  be  used  in 
Arithmetic  ;  and  the  Arabic  figures  are  used  in  all  de^ 
partments  of  mathematics.  This  method  of  represent- 
ing quantities  by  letters  is  often  called  the  Algebraic 
method,  and  the  method  b}'-  the  Arabic  characters,  the 
Arithmetical.  It  would  be  better  to  call  the  former  the 
Literal  method,  and  the  latter  the  Decimal. 

This  part  [Part  I.  ]  of  this  volume  is  nothing  but  a  chapter  upon 
Arithmetic,  showing  the  principles  of  the  Literal  Notation,  and  how 
the  several  Arithmetical  operations  of  addition,  subtraction,  multipli- 
cation, etc. ,  are  performed  by  means  of  this  notation. 

28,  Tlie  Literal  If otationh^^  some  very  great  ad- 
vantages over  the  decimal  for  purposes  of  mathemati- 
cal reasoning.  1st,  The  symbols  are  more  general  in 
their  signification  ;  and  2d,  We  are  enabled  to  detect 
the  same  quantity  anywhere  in  the  process,  and  even  in 
the  result.  Thus  it  happens  that  the  processes  become 
general  formuloe,  or  rules,  instead  of  special  solutions. 

IXiii. — These  statements  are  very  important,  but  it  is  questionable 
•whether  young  beginners  can  be  made  to  comprehend  them  fully  ;  and 
all  will  see  their  meaning  better  after  having  studied  Algebra  some 
time.  But  we  will  try  and  make  them  as  plain  as  possible  now.  To 
illustrate  the  first  statement,  suppose  we  say  a  boy  has  7  apples  ;  you 
know  just  how  many  are  meant.  But  when  we  say  a  boy  has  h  ap- 
ples, nobody  can  tell  how  many  he  has.  In  fact,  it  is  not  designed  to 
teU  the  exact  number,  but  only  to  say  that  he  has  some  number. 
Again,  7  represents  the  same  number  of  units  always  ;  but  a  letter 
may  be  used  to  represent  any  number  of  units  we  please  ;  or,  it  may 
be  used,  as  we  have  just  said,  without  our  caring  to  specify  any  pre- 
cise number  of  units.  This  may  seem  to  be  a  very  unsatis- 
factory kind  of  notation ;  but  with  patience  its  advantages  voll  appear. 
Consider  the  following  examples  : 


10  FUNDAMENTAL    RULES. 


EXAMPLES. 

1.  A  boy  has  8  marbles  which  he  sells  for  three  cents  each,  and 
takes  his  pay  in  pencils  at  6  cents  each.  How  many  pen- 
cils does  he  receive  ?     Suppose  we  answer  that  he  receives 

3  times  8  divided  by  6,  or  pencils,  without  giving 

the  number  more  explicitly.  .  .  Now  take  a  similar 
example,  using  the  literal  notation  :  thus,  A  boy  sells  a 
certain  number  of  marbles  which  we  will  represent  by  c, 
for  a  number  of  cents  each,  which  we  will  call  m,  and 
takes  his  pay  in  pencils  at  b  cents  each.  How  many  pen- 
cils does  he  receive  ?     We  will  answer  as  before,  and  say 

he  receives  c  times  m,  divided  by  h,  or  — - —  pencils.  The 

pupil  will  notice  this  difference  between  the  answers  ; 
both,  as  they  now  stand,  simply  tell  what  operations  to 
perform  in  order  to  get  the  answers  ;  but,  in  the  former 
case,  we  can  perform  the  operations  and  get  the  explicit 
answer,  4,  while  in  the  latter  case,  we  can  only  leave  it 

C  X  771 

as  it  is.  Such  an  answer  as  — j—  may  seem  to  the  pu- 
pil to  be  no  answer  at  all ;  and  indeed  it  is  not  an  an- 
swer in  the  same  sense  as  he  has  been  accustomed  to 
think  of  answers  ;  nevertheless  it  is  often  more  useful. 
Notice  that  the  answer  4  is  only  true  for  the  specific  ex- 

C  X  71% 

ample,  while  the  answer  — —  is  true  in  every  like  exam^ 

pie.  No  matter  what  the  number  of  marbles,  what  the 
price,  or  what  the  price  of  a  pencil,  a  general  answer  is 

C  X  772- 

— T — .     We  also  observe  that  the  quantities  3,  8,  and  6 

do  not  appear  distinctly  in  the  answer  4  ;  but  the  c,  m, 
and  h  do,  and  would,  in  general,  however  complicated 
the  problem. 


NOTATION.  11 

2.  One  boy  sold  5  pears  at  3  cents  each  ;  another  sold  6 
apples  at  2  cents  each  ;  and  a  third  sold  3  melons  at  8 
cents  each.     How  much  did  they  all  receive  ? 

Ans.  51  cents. 

3.  One  boy  sold  b  pears  at  c  cents  each  ;  another  sold  m 
apples  at  n  cents  each ;  a  third  sold  d  melons  at  g  cents 
each.     How  much  did  they  all  receive  ? 

Ans.  bXG-{-  mxn -{-dxg  cents. 

Suggestions. — Notice  that  in  the  3d  example  the  several  quantities 
of  the  problem  are  distinctly  seen  in  the  answer,  but  not  so  in  the 
answer  to  M.  2.  Moreover,  the  answer  to  Ex.  3  is  equally  true  for 
any  and  all  values  of  b,  c,  m,  n,  d  and  g.  Consider  in  like  manner 
the  two  following  : 

4.  If  I  buy  5  cords  of  wood  at  4  dollars  per  cord,  and  pay 
for  it  in  cloth  at  2  dollars  jDer  yard,  how  many  yards 
are  required?  Aus.  10  yards. 

5.  If  I  buy  a  cords  of  wood  at  b  dollars  per  cord,  and  pay 

for  it  in  cloth  at  c  dollars  per  yard,  how  many  yards  are 

required?  ,  axb 

Ans.     yards. 

6.  A  man  had  a  flock  of  m  sheep.  He  lost  2n  of  them 
and  raised  10a.  After  which  he  sold  the  flock  at  $c  per 
head,  taking  his  payment  in  cloth  at  $6  per  yard. 
What  operations  must  be  performed  on  these  numbers 
in  order  to  ascertain  the  number  of  yards  of  cloth  re- 
ceived?    And  how  will  this  number  be  expressed  ? 

c(m— 2>i+10a) 

Ans.  ■ r * 

b 

7.  A  man  bought  3  horses.  He  gave  for  the  first  twice  as 
much  as  for  the  second,  and  for  the  third  c  times  as 
much  as  for  both  the  others.  If  x  represents  the  price 
of  the  first,  how  much  did  he  give  for  the  3d  ? 

Ans.   {x-\--)c. 


12  FUNDAMENTAL   RULES. 

20^  In  using*  the  decimal  notation  certain  laws  are  es- 
tablished in  accordance  with  which  all  numbers  can  be  rep- 
resented by  the  ten  figures.  Thus,  it  is  agTeed  that  when 
several  figures  stand  together  without  any  other  mark,  as 
435,  the  right  hand  figure  shall  signify  units,  the  second  to 
the  left,  tens,  the  third,  hundreds,  etc. ;  also  that  the  sum 
of  the  several  values  shall  be  taken.  This  number  is, 
therefore,  4  hundreds  +  3  tens  +  5  (units). 

SO,  In  hke  manner,  certain  laws  are  observed  in  repre- 
senting numbers  by  letters. 

FiKST  Law. 

Known  Quantities,  that  is  such  as  are  given  in  a 
problem,  are  represented  by  letters  taken  from  the  first 
part  of  the  alphabet ;  while  XTnlznoivn  Quantities, 
or  quantities  whose  values  are  to  be  found,  are  represented 
by  letters  taken  from  the  latter  part  of  the  alphabet. 

III. — A  grocer  has  two  kinds  of  tea,  one  of  which  is  worth  a  cents 
(any  qivtn  number  being  meant  by  a)  j)er  pound,  and  the  other  h 
cents.  Row  many  pounds  of  each  must  he  take  to  make  a  chest  of  e 
pounds,  which  will  be  worth  d  dollars  ?  In  this  problem,  a,  h,  c,  and 
d  are  the  given  or  known  quantities,  and  hence  are  represented  by 
letters  from  the  first  part  of  the  alphabet.  The  unknown  or  required 
quantities  are  the  number  of  pounds  of  each  of  the  two  kinds  of  tea. 
We  therefore  represent  the  number  of  pounds  of  the  first  kind  by  ic, 
and  of  the  second  kind  by  y. 

ScH. — This  law  is  not  very"  rigidly  adhered  to,  except  that  letters 
after  and  including  v,  are  generally  used  to  represent  unknown  quan- 
tities, while  the  others  are  used  for  known  quantities.  But  it  is  some- 
times convenient  to  use  a  different  notation.  Thus,  in  problems  in 
Interest,  the  principal  may  be  represented  by  p,  whether  it  is  known 
or  unknown,  the  interest,  in  like  manner,  by  i,  the  rate  per  cent,  by 
r,  the  time  by  t,  etc. 

Accented  letters,  as  a',  a",  a"\  a"",  &c.,  (read  "a  prime,"  "a 
second,"  "a  third,"  etc.,)  and  letters  with  subscripts,  aso-i.,  a^, 


NOTATION.  IS 

Cj,  ^4,  etc.,  (read  "a  sub  1,"  ''a sub  two,"  etc.,)  are  sometime^ 
used.  This  form  of  notation  is  used  when  there  are  sev- 
eral like  quantities  in  the  same  problem,  but  which  have 
different  numerical  values.  Thus,  in  a  problem  in  vv^hich 
several  walls  of  different  heights,  breadths,  and  lengths,  are 
considered,  we  may  represent  the  several  heights  by  a\ 
c."  a",  etc.,  or  a^,  a..,  a„  etc.  ;  'the  thicknesses  by  h',  h",. 
l'\  etc.,  or  bi,  b.,  h-,  etc.,  and  the  lengths  by  l',  I",  I'", 
etc.,  or  /i,  I2,  k,  etc. 

The  Greek  letters  are  also  often  used  both  for  known 
and  unknown  quantities. 

The  student  will  notice  a  difference  between  Algebra 
and  Arithmetic,  in  that,  in  Algebra,  the  unknown  quanti- 
ties (what  he  has  called  the  "Answers"  in  Arithmetic) 
are  represented  in  solving  a  problem,  and  are  used  in  the 
solution  just  like  known  quantities.  This  device  gives 
Algebra  a  great  advantage  over  Arithmetic. 

Second  Law. 
"When   letters   are  written   in   connection,  without  any 
sign   between   them,  their  product  is  signified.     Thus  abo 
signifies  that  the  three  numbers  represented  by  a,  b,  and  c 
are  to  be  multiphed  together. 

ScH.  1.  — There  is  here  an  interesting  difference  between  this  nota- 
tion and  the  decimal.  There  are  two  points  of  difference  ;  1st,  The 
place  of  a  letter,  as  at  the  right  or  left,  has  nothing  to  do  with  its 
value  ;  and  2d,  The  sign  understood  between  them  is  that  of  multi- 
plication instead  of  addition,  as  in  the  decimal  notation.  In  the  deci- 
mal system,  if  we  write  the  three  figures  5,  4,  and  3,  as  we  have 
v»Titten  the  letters  a,  h,  and  c,  thus  5i3  ;  1st,  Each  figure  has  a  par- 
ticular value  dependent  upon  the  place  it  occupies,  the  5  representing 
.  hundreds,  the  4  tens,  and  the  3  units ;  2d,  The  amoiint  represented  i^ 
500  -f-  40  -f-  3.  Moreover,  the  several  le!:ters  are  not  at  all  restricted 
in  signification  ;  «  may  represent  5,  48,  10.06,  or  any  other  number, 
however  small  or  however  large,  and  integral,  fractional,  or  mixed  ; 
and  the  same  is  true  of  any  other  letter.  In  fact,  the  meaning  usually 
is,  that  a  represents  any  number,  h  any  other,  and  c  any  other,  etc 


1  J:  FUNDAMENTAL  EULES. 

The  same  letter,  however,  means  the  same  thing  throughout  one 
problem.  Such  expressions  as  abc,  mnxy,  etc.,  are  read  by  simply 
naming  the  letters  in  order. 

ScH.  2. — When  figures  are  written  with  letters,  their  relation  to  the 
letters  is  the  same  as  that  of  the  letters  to  each  other.  Thus  4a6 
means  the  continued  product  of  4,  a,  and  h.  Also  125xy  means  the 
continued  product  of  125,  x,  and  y. 

31,  A  character  like  a  figure  8  placed  horizontally,  oo , 
is  used  to  represent  what  is  called  Infinity^  or  a  quantity 
larger  than  any  assignable  quantity. 

ScH. — As  the  mathematical  notion  of  infinity  is  always  a  trouble- 
some one  to  learners,  we  shall  lose  no  opportunity  for  making  it  clear. 
By  an  infinite  quantity  is  not  meant  one  larger  than  any  other,  or  the 
largest  possible  quantity.  It  simply  means  a  quantity  larger  than  any 
assignable  quantity  ;  i.  e.,  larger  than  any  one  which  has  limits.  The 
mathematical  notion  concerns  rather  the  manner  of  conceiving  the 
quantity,  than  its  absolute  value.  Thus,  a  series  of  Is,  as  1  1  1,  etc., 
repeated  without  stopping,  represents  an  infinite  quantity,  because, 
from  the  method  of  conceiving  the  quantity,  it  is  necessarily  greater 
than  any  quantity  which  we  can  assign  or  mention.  If  we  assign  a 
row  of  9s  reaching  around  the  world,  though  it  is  an  inconceivably 
great  number,  it  is  not  as  great  as  a  series  of  Is  extending  without  limit. 
Moreover,  one  infinite  may  be  larger  than  another  ;  for  a  series  of  2s 
extending  without  limit,  as  2  2  2  2,  etc.,  is  twice  as  large  as  a  series  of 
Is  conceived  in  the  same  way.  It  is  never  of  any  use  to  try  to  com- 
prehend the  magnitude  of  an  infinite  quantity  ;  we  can  not  do  it  ;  al- 
though we  can  compare  infinites  just  as  well  as  finites.  It  will  be 
necessary  to  say  much  more  on  this  interesting  and  important,  though 
somewhat  puzzling  subject,  farther  on  in  our  course. 

SYMBOLS     OF     OPERATION. 

32,  The  Symbols  of  Operation  used  in  Algebra 
are  the  same  as  in  Arithmetic,  or  in  any  other  branch  of 
mathematics  ;  but  to  refresh  the  memory,  we  will  repeat 
them. 

33,  The  perpendicular  cross,  +,  is  called  the  plus  sign, 
and  read  "  i)lus."     It  signifies  that  the  quantities  between 


NOTATION.  ■        15 

which  it  is  placed  are  added  together.    Thus,  a  +  2c6  -f-  xm 
is  read,  "  a  plus  2cb  plus  xm." 

34,  A  short  horizontal  line,  — ,  is  called  the  minus  sign, 
and  is  read  "  minus."  It  signifies  that  the  quantity  before 
which  it  is  placed  is  to  be  subtracted.  Thus  a  —  2cb  —  xm 
+  12ax  is  read,  "  a  minus  2cb  minus  xm  plus  12ax" 

35,  An  S-shaped  symbol  placed  horizontally,  ^^,  is 
sometimes  used  to  signify  the  difference  between  two  quan- 
tities, a  ^v^  6  is  read,  "  the  difference  between  a  and  6."  This 
sign  differs  from  the  preceding  in  that  it  does  not  indicate 
which  of  the  two  quantities  is  to  be  taken  as  the  subtra- 
hend, while  the  minus  sign  requires  us  to  consider  tha 
quantity  before  which  it  is  placed  as  the  subtrahend. 

30,  The  obHque  cress,  X,  and  a  simple  dot,  • ,  are  each 
signs  of  multiplication.  In  the  case  of  literal  factors,  the 
sign  is  usually  omitted,  according  to  the  second  law  of 
notation.  Thus,  4  X  ^  X  ^>  ^"^'^j  and  ^^ac,  signify  exactly 
the  same  thing. 

37,  The  signs  of  division  are,  a  horizontal  hne  between 

two  dots,  having  the  dividend  at  the  left  and  the  divisor  at 

the  right,  as  12ac-i-25;  or  the  dots  without  the  line,  as  12ao 

:  2b  ;  or  the  line  without  the  dots,  the  dividend  being  writ- 

12ac 
ten  above  and  the  di\dsor  below  it,  as  ^r  ;  all  of  which 

2b 

are  read,  "  12ac  divided  by  2b."  In  performing  division,  the 
divisor  is  sometimes  written  at  the  left  of  the  dividend  and 
separated  from  it  by  a  curved  hne  ;  the  quotient  is  then 
written  at  the  right  and  separated  from  the  dividend  in  the 
same  manner  :  as,  2a)12ac{Gc,  in  which  2a  is  the  divisor, 
12ac  the  dividend,  and  6c  the  quotient.  Sometimes,  espe- 
cially in  Algebra,  the  divisor  is  written  on  the  right  of  the 
dividend,  and  separated  from  it  by  a  vertical  line,  the  quo- 


IG  FUNDAMENTAL   RULES, 

tient  in  tliis  case  being  written  under  the  divisor.     Thus, 
12ac  I  2a  is  the  last  example  above,  expressed  in  a  diiler- 

6c 
ent  form. 

ScH. — It  is  very  inelegant,  though  quite  common  in  some  j)arts  of 

tlie  country,  to  read  such  expressions  as  — — — ,  "  12ac  over  25."      Wa 

should  read  "  12ac  divided  by  2^." 

[Note. — Let  great  care  be  taken  that  the  nature  of  exponents,  as  ex- 
plained in  the  succeeding  articles,  be  clearly  comprehended.  No  lit- 
tle difficulty  arises  from  an  imperfect  understanding  of  this  notation. 
A  very  common,  though  very  erroneous  method  of  reading  such  ex- 
pressions, greatly  aggravates  the  difficulty.  ] 

38,  A  JPower  of  a  number  is  the  product  which 
arises  from  multiplying  the  number  by  itself,  i.  e.,  taking  it 
a  certain  number  of  times  as  a  factor. 

30,  A.  Moot  of  a  number  is  one  of  several  equal  fac- 
tors into  which  the  number  is  to  be  resolved. 

40,  An  Exponent  is  a  small  figure,  letter  or  other 
symbol  of  number,  written  at  the  right  and  a  little  above 
another  figure,  letter  or  symbol  of  number. 

41,  A  JPositive  Integral  Exponent  signifies 
that  the  number  afi'ected  by  ic  is  to  be  taken  as  a  factor  as 
many  times  as  there  are  units  in  the  exponent.  It  is  a 
kind  of  symbol  of  multiphcation. 

III.  2^,  read,  "  2,  third  power,"  signifies  that  two  is  to  bo  taken 
three  times  as  a  factor,  i.  e.,  2  X  2  X  2,  or  8.  3*,  read,  "  3,  fourth 
power,"  signifies  3  X  3  X  3  X  3,  or  81  ;  81  is  the  fourth  power  of 
3,  because  it  is  the  product  of  3  taken  four  times  as  a  factor,  a^  is 
aaaaa.  x*",  read  "a,  iwth power,"  or  "x,  exponent  m,"  is. xajx.  .  .etc., 
till  m  factors  of  x  are  taken. 

42,  A  Positive  Fractional  Bxponent  indi- 
cates a  power  of  a  root,  or  a  root  of  a  power.  The  de- 
nominator specifies  the  root,  and  the  numerator  the  power 
of  the  number  to  which  the  exponent  is  attached. 


DOTATION.  17 

III.  8^,  read,  "  8,  exponent  f,"  not  "  8,  |  power,"  (there  is  no  such 
tiling  as  a  f  power)  is  the  second  power  of  the  third  root  of  8.     The 

third  root  of  8  being  2,  and  the  second  power  of  2  being  4,  8=*  =  4. 
"We  may  also  understand  the  power  to  be  taken  first,  and  then  the 

3. 

root,  as  will  be  demonstrated  hereafter.  Thus,  8'*  is  the  third  root  of 
the  second  power  of  8.  The  second  power  of  8  is  64,  the  third  root 
of  which  is  4,  which  is  the  same  result  as  was  obtained  by  taking  the 

root  first,  and  then  the  power.     (125)  ^  is  5.     (125)  ^  is  25.     (32)  ^  is 

m 

8.  £c»  ,  read,  "x,  exponent  jn  divided  by  ??,"  means  that  x  is  to  be  re- 
solved into  n  equal  factors,  and  the  product  of  m  such  factors  taken. 

43.  A  Negative  Ex2)onent^  i.  e.,  one  with  tlio  — 
sign  before  it, — either  integral  or  fractional,  signifies  the 
reciprocal  of  what  the  expression  would  be  if  the  exponent 
were  positive, — i.  e.,  had  tho  +  sign,  or  no  sign  at  all  be- 
fore it. 

-4  1  1  _3.     1 

III.     3      ,  read  "  3,  exponent  — 4,"  signifies  -^>    or  ttt"    2       ^^~^^' 

1  1  -2  1  1  _:i  .      1 

or  -rr  •     ar-**  is  — -    etc.     Also  8    "^  is    — ;'  or  —r '     m      ''is  — « • 

44,  The  Radical  Sign^  V,  is  also  used  to  indicate 
the  square  root  of  a  quantity.  When  any  other  than  the 
square  root  is  to  be  designated  by  this,  a  small  figure  speci- 
fying the  root  is  placed  in  the  sign.     Thus  v^  5  signifies 

the  3d,  or  cube  root  of  5,  and  is  the  same  as  5^.      v/~34ar^' 

indicates  the  5th  root  of  34a63  and  is  the  same  as  (34a63)^. 


ScH. — Read  "v/oac,  "the  square  root  of  5ac,"  not  "radical  Sac' 
The  latter  expression  is  generic,  and  appUes  as  well  to  F  5ac,  or  \/dac. 
Besides  it  is  inelegant. 

Let  the  pupil  read  the  following  examples  and  give  the 
signification  of  each. 


J^' 


18  FUNDAMENTAL   RULES. 


EXAMPLES. 
2 

1.  25a-26^.  Bead,  "25,  a  exponent  — 2,  6  exponent  f."  It 
means  25  multiplied  by  — ,  multiplied  by  the  square 
of  the  cube  root  of  b. 

2.  ar^~^.    Kead,  "jt,  exponent -1."     Since  —   —  1   is 

n  n 

,  X  »      is  the  same  as  a;   "   ,  and  hence  means  that 

n 

X  is  to  be  raised  to  a  power  indicated  by  m  —  n,  and 

the  nth  root  of  this  power  extracted. 

3.  Bead  and  explain  ^d^h~''\     12x"'y  ~  ~. 

SYMBOLS    OF    RELATION. 

45.  The  Sign  of  Geometrical  Matio  is  two 

dots  in  the  form  of  a  colon,  :  .  Thus  a  :  b,  is  read  "  a  is  to 
b,"  or,  "  the  ratio  of  a  to  6."     It  means  the  same  as  a-^b. 

46.  The  Sign  of  Aritlimetical  Hatio  is  two 

dots  placed  horizontally,  ••  •  Thus  a  ••  6  is  read, 
"  the  Arithmetical  ratio  of  a  to  6  and  is  equivalent  to  a — b. 

47.  The  Sign  of  Equality  is  two  parallel  hori- 
zontal hnes,  =.  Thas,  2cj;  =^  xy,  is  read,  "  2gx  equals  xy" 
5ac  —  2by  ==  dx-  is  read,  "  5ac  minus  2by  equals  3x^." 

I  Four  dots  in  the  form  of  a  double  colon,  :  :,  is  the  sign 
of  equality  between  ratios.  Thus,  a  :  b  :  :  c  :  d,  read,  "  a  is 
to  6  as  c  is  to  d,"  means  that  the  ratio  of  a  to  b  equals  the 
ratio  of  c  to  d,  and  may  just  as  well  be  written  a  :b  =  c  :  d, 

CL  C 

or  r  =  -,  all  of  which  expressions  mean  exactly  the  samo 
0       d 

thing. 

48.  The  Sign  of  Inequality  is  a  character 
somewhat  like  a  capital  V  placed  on  its  side,  <^,  the  open- 


NOTATION.  19 

ing  being  towards  the  greater  quantity.  Thus  a'^b  is 
read,  "a  greater  than  6."  m<^n  is  read,  "  m  is  less  than 
n." 

49.  Tlie  Sign  of  Variation  is  somewhat  hke  a 

figure  8  open  at  one  end  and  placed  horizontally.     Thus, 

c  c 

fl  CXI   J  is  read,  "  a  varies  as  -." 
a  d 


k 


SYMBOLS   OF   AGGREGATION. 


50.  A  Vinculum  is  a  horizontal  line  placed  over 
several  terms,  and  indicates  that  they  are  to  be  taken  to- 
gether.    The  parenthesis,  (  ),  the  brackets,   [  ],  and  the 

brace,   \   \ ,  have  the  same  sismification. 


!!■' 


Ill,  a  -^  by^cd  —  e  means  that  (a  -|-  &)  is  to  be  multiplied  by 
(cd  —  e).  (a  -j-  &)  X  (c^  —  ^)  ^^'^  means  the  product  of  (a  -j-  &)  and 
(cd  —  e).     Brackets  and  braces  are  used  when  one  parenthesis  would 

fall  within  another.    Thus,     ]  2  +  [a  +  (5  +  c)x]?/  i  u,  signifies  that 

the  product  of  (6  -|-  c)  multiplied  by  x,  is  to  be  added  to  a,  and  this 
sum  multiphed  by  y ;  to  this  product  z  is  to  be  added  and  the  sum 
multiplied  by  u. 

51.  A  vertical  line  after  a  column  of  quantities,  each 
having  its  own  sign,  signifies  that  the  aggregate  of  the  col- 


umn is  to  be  taken  as  one  quantity.     Thus  -\-a 
same  as  (a  —  6  +  c)x.  — h 


X  is  the 


SYMBOLS   OF   CONTINUATION. 

S2*  A  series  of  dots, ,  or  of  short  dashes, 

,  written  after  a  series  of  expressions,  signifies  "  &c." 

Thus  a :  ar :  ar^ :  ar^ ar"  means  that  the  series  is 

to  be  extended  from  ar^  to  ar»  ,  whatever  may  be  the  value 
of  n. 


20  FUNDAMENTAL   EULES. 


SYMBOLS    OF    DEDUCTION. 

SS,  Three  dots,  two  being  placed  lK)rizontally  and  the 
third  above  and  between,  .  • .  ,  signify  therefore,  or  some  ana- 
logous expression.  If  the  third  dot  is  below  the  first  two, 
• .  • ,  the  symbol  is  read  "  since,"  "  because,"  or  by  some 
t  ]uivalent  expression. 

POSITIVE   AND    NEGATIVE    QUANTITIES. 

54,  Positive  and  Negative  are  terms  primarily 
applied  to  concrete  quantities  which  are,  by  the  conditions 
of  a  problem,  opposed  in  character. 

III. — A  man's  property  may  be  called  positive,  and  his  debts  nega- 
tive. Distance  up  may  be  called  positive,  and  distance  down,  nega- 
tive. Time  before  a  given  period  may  be  called  positive,  and  aftei^, 
negative.  Degrees  above  0  on  the  thermometer  scale  are  called  posi- 
tive, and  below,  negative. 

55.  The  signs  +  and  —  are  used  to  indicate  the  char- 
acter of  quantities  as  positive  or  negative,  as  well  as  for  the 
purpose  of  indicating  addition  and  subtraction. 

This  double  signification  of  the  signs  -(-  and  —  gives  rise  to  some 
confusion,  to  avoid  which  we  shall  in  our  elementary  illustrations  use 
the  small  hght  lined  signs,  +  - ,  to  mark  the  distinction  of  positive 
and  negative,  and  the  common  sized  characters,  -| ,  to  signify  addi- 
tion and  subtraction.  After  the  principle  has  become  familiarized, 
this  distinction  will  be  dropped,  as  it  is  not  used  in  practice. 

SO,  In  problems  in  which  the  distinction  of  positive  and 
negative  is  made,  each  quantity  in  the  formidce  is  to  be  con- 
sidered as  having  a  sign  of  character  expressed  or  under- 
stood besides  the  plus  or  minus  sign,  which  latter  indicates 
that  it  is  to  be  added  or  subtracted.  The  positive  sign  need 
not  be  written  to  indicate  character,  as  it  is  customary  to 
consider  quantities  whose  character  is  not  specified  as 
positive. 


NOTATION.  21 

III.  1. — In  the  expression  at;  -f  m  —  ex,  let  the  problem  out  of  which 
it  arose  be  such,  that  a,  m,  and  x,  tend  to  a  positive  result,  and  h  and  c 
to  an  opposite,  or  a  negative  result.  Giving  these  quantities  theii-  signs 
of  character,  we  have,  (  -t-a)  X  ir^)  +  (  ■*-»^)  —  (-c)  X  (  ■^^^)  which  may 
be  read,  "positive  a  multiplied  by  negative  &,  plus  positive  m,  minus 
negative  c  multiplied  by  positive  x. "  Suppressing  the  positive  sign 
this  maybe  written,  a{-b)  +  m  —  (-c)x,  by  also  omitting  the  unneces. 
sary  sign  of  multiplication. 

III.  2. — As  this  subject  is  one  of  fundamental  importance,  let  carefu) 
attention  be  given  to  some  further  illustrations.  We  are  to  distinguish 
between  discussions  of  the  relations  between  mere  abstract  quantities, 
and  problems  in  which  the  quantities  have  some  concrete  signification. 
Thus,  if  it  is  desired  to  ascertain  the  sum  or  difference  of  468,  or  m, 
and  327,  or  n,  as  mere  numbers,  the  question  is  one  concerning  tha 
relation  of  abstract  numbers,  or  quantities.  No  other  idea  is  attached 
to  the  expressions  than  that  they  represent  a  certain  number  of  units. 
But,  if  we  ask  how  far  a  man  is  from  his  starting  point,  who  has  gone, 
first,  468,  or  m  miles  directly  east,  and  then  327,  or  n  miles  directly 
west ;  or  if  we  ask  what  is  the  difference  in  time  between  468,  or  m 
years  B.  C,  and  327,  or  n  years  A.  D.,  the  numbers  468,  or  m,  and 
327,  or  n,  take  on,  besides  their  primary  signification  as  quantities, 
the  additional  thought  of  opposition  in  direction.  They  therefore  be- 
come, in  this  sense,  concrete. 

Again,  a  company  of  5  boys  are  trying  to  move  a  wagon.  Three  of 
the  boys  can  puU  75,  85,  and  100  pounds  each  ;  and  they  exert  their 
strength  to  move  the  wagon  east.  The  other  two  boys  can  pull  90  and 
110  pounds  each  ;  and  they  exert  their  strength  to  move  the  wagon 
west.  It  is  evident  that  the  75,  85,  and  100  are  quantities  of  an  oppo- 
site character,  in  their  relation  to  the  problem,  from  90  and  110.  Again, 
suppose  a  party  rowing  a  boat  up  a  river.  Their  united  strength 
would  propel  the  boat  8  miles  per  hour  if  there  were  no  cun-ent ;  but 
the  force  of  the  current  is  sufficient  to  carry  the  boat  2  miles  pei 
hour.  The  8  and  2  are  quantities  of  opposite  character  in  their  rela- 
tion to  the  problem.  Once  more,  in  examining  into  a  man's  business, 
it  is  found  that  he  has  a  farm  worth  m  dollars,  personal  property  worth 
n  dollars,  and  accounts  due  him  worth  c  dollars.  There  is  a  mortgage 
on  his  farm  of  h  dollars,  and  he  owes  en  account  a  dollars.  The  m,  n, 
and  c  are  quantities  opposite  in  their  nature  to  h  and  cX.  This  opposi- 
tion in  character  is  indicated  by  calling  those  quantities  which  contribute  to 
one  result  positive,  and  those  which  contribute  to  the  opposite  result  ne^Or- 
iive. 


22  FUNDAMENTAL   RULES. 

57.  Purely  abstract  quantities  have,  properly,  no  dis- 
tinction as  positive  and  negative  ;  but,  since  in  such  prob- 
lems the  plus  or  additive,  and  the  minus  or  subtractive 
terms  stand  in  the  same  relation  to  each  other  as  positive 
and  negative  quantities,  it  is  customary  to  call  them  such. 

III. — In  the  expression,  5ac  —  Scd  -\-  8xy  —  2ad,  though  the  quan- 
tities, a,  c,  d,  X  and  y  be  merely  abstract,  and  have  no  proper  signs  of 
character  of  their  own,  the  terrns  do  stand  in  the  same  relation  to  each 
other  and  to  the  result,  as  do  positive  and  negative  quantities.  Thus, 
Sac  and  8xy  tend,  as  we  may  say,  to  increase  the  result,  while  —  Scd, 
and  —  2ad  tend  to  diminish  it.  Therefore  the  former  may  be  called 
positive  terms,  and  the  latter  negative. 

oSm  ScH. — Less  than  zero.  Negative  quantities  are  frequently  spoken 
of  as  ' '  less  than  zero. "  Though  this  language  is  not  philosophically 
correct,  it  is  in  such  common  use,  and  the  thing  signified  is  so  sharply 
defined  and  easily  comprehended,  that  it  may  possibly  be  allowed  as 
a  conventionahsm.  To  illustrate  its  meaning,  suppose  in  speaking  of 
a  man's  pecuniary  afiairs  it  is  said  that  he  is  worth  "less  than  noth- 
ing ;"  it  is  simply  meant  that  his  debts  exceed  his  assets.  K  this 
excess  were  $1000,  it  might  be  called  negative  $1000,  or  -$1000.  So, 
again,  if  a  man  were  attempting  to  row  a  boat  up  a  stream,  but  with 
all  his  effort  the  current  bore  him  down,  his  progress  might  be  said  to 
be  less  than  nothing,  or  negative.  In  short,  in  any  case  where  quan- 
tities are  reckoned  both  ways  from  zero,  if  we  call  those  reckoned  one 
way  greater  than  zero,  or  positive,  we  may  call  those  reckoned  the 
other  way  "less  than  zero,"  or  negative. 

50,  The  value  of  a  Negative  Quantity  is  conceived  to 
increase  as  its  numerical  value  decreases. 

III. — Thus  -3  >>  -5,  as  a  man  who  is  in  debt  $3,  is  better  off 
than  one  who  is  in  debt  $5,  other  things  being  equal.  If  a  man  is  striv- 
ing to  row  up  stream,  and  at  first  is  borne  down  5  miles  an  hour,  but 
by  practice  comes  to  row  so  well  as  only  to  be  borne  down  3  miles  an 
hour,  he  is  evidently  gaining  ;  i.  e.,  -3  is  an  increase  upon  -5. 
Finally,  consider  the  thermometer  scale.  K  the  mercury  stands  at  20^ 
below  0,  (marked  -20^)  at  one  hour,  and  at  -10°  the  next  hour, 
the  temperature  is  increasing  ;  and,  if  it  increase  sufficiently  will  b^ 
come  0,  passing  which  it  will  reach   -+■  1°,    -*•  2°,  etc.     In  this  illustra- 


MOTATION.  /  23 

Hon,  the  quantity  passes  from  negative  to  positive  by  passing  through  0. 
This  is  assumed  as  a  fundamental  truth  of  the  doctrine  of  positive  and 
negative  quantities,   viz.  :    That  a  quantity  in  passing  through  0 

CHANGES   ITS   SIGN. 

It  appears  in  geometrj^,  that  a  quantity  may  also  change  its  sign  in 
passing  through  infinity.  Thus  the  tangent  of  an  arc  less  than  90^  is 
positive  ;  but  if  the  aic  continually  increases,  the  tangent  becomes  in- 
finity at  90  ,  passing  which  it  becomes  negative.  From  this  illustra- 
tion it  also  appears  that  negative  infinity,  -co,  is,  sometimes,  at 
least,  to  be  considered  as  an  increase  upon  positive  infinity,  -»-  (x. 
The  importance  of  these  considerations  appears  in  Trigonometry  and 
especially  in  the  Calculus. 


NAMES  OF  DIFFERENT  F0R3IS  OF  EXPRESSION. 

00*  A.  I*olynoinial  is  an  expression  composed  of 
two  or  more  parts  connected  by  the  signs  plus  and  minus, 
each  of  which  parts  is  called  a  term. 

01*  A-  3£oilomial   is    an    expression    consisting    of 

one  term  ;  a  Sinoniial  has  two  terms;  a  Trinomial 

has  three  terms  ;  etc. 

1 
III.     5a-b  —  cd  '^  -{-  x  —  4  (a  +  &)  is  a  polynomial  of  4  terms.     The 


first  three  arc  monomial  terms,  and  the  last  is  a  binomial  term.     5ac 
—  ef  and  x-  + 
is  a  trinomial. 


ef  and  x-  +  y-  are  examples  of  binomials.     2a^x'^y  —  125d—^  -\-  12 


02,  A  Coefficient  of  a  term  is  that  factor  which  is 
considered  as  denoting  the  number  of  times  the  remainder 
of  the  term  is  taken.  The  numerical  factor,  or  the  pro- 
duct of  the  known  factors  in  a  term  is  most  commonly 
called  the  coefficient,  though  any  factor,  or  the  product  of 
any  number  of  factors  in  a  term  may  be  considered  as 
coefficient  to  the  other  part  of  the  term. 

III. — In  the  term  6a,  G  is  the  coefficient  of  a.  In  ax,  a  may  be 
called  the  coefficient  of  x,  or  1  may  be  called  the  coefficient  of  ax.  In 
6axj/,  6  is  the  coefficient  of  axy,  6a  oixy,  and  &ax  of  y.  In  5,a6  —  c\ 
5  is  the  coefficient  of  {ab  —  c)  ;  and  in  {2a^  —  cd)  xy,  i2a'  —  cd)  ia 
the  coefficient  of  xy. 


24  FUNDAMENTAL   EULES.  . 

03,  Similar  Terms  are  sucla  as  consist  of  the  same 
letters  affected  with  the  same  exponents. 

luD.     5ab,  13db,  and  ah  are  similar.     —  IGx^-ip,  5x'y^,&nd  —  jc*? 
are  also  similar  to  each  other.    4a&,  Bah-,  and  —  2ab^  are  dissimilar,  as 

are  Sax,  —  56a;^,  4:cx%  and  5xy. 


EXERCISES  IN  NOTATION. 

1.  "Write  in  mathematical  symbols,  5  times  the  square  root 
of  a,  added  to  the  cube  of  the  sum  of  a  and  b. 

Result,  (a  +  by  +  Wa,  or  {a  +  by  +  5a^. 

How  many  terms  in  this  result  ?     What  kind  of  a  term 
is  the  first  one  ? 

2.  Write  the  second  power  of  a,  plus  3  times  the  product 
of  c  square  multiplied  by  b,  diminished  by  m  times  the 
cube  root  of  the  binomial,  the  square  root  of  a  minus 
the  cube  of  b. 


Result,  a-  +  dc-b  —  m  (a-  —  ^3)  ^,  or  a-  +  ^c^b  — m^a^^  —  6\ 

3.  Write  three  times  a  into  5,  plus  the  binomial  a  minus  6, 
divided  by  the  sum  of  a  square  and  b  cube. 

4.  Write  the  fraction,  the  product  of  the  sum  of  a  and  b 
into  the  sum  of  x  and  \j,  divided  by  the  square  root  of  a 
diminished  by  the  cube  root  of  b. 

j^esult,  (-  +  ^)(-^y). 
^a  —</b 

5.  Write  the  fraction,  a  fifth  power  diminished  by  3  times 
a  square  b  cube,  divided  by  the  square  root  of  the  bi- 
nomial X  square  diminished  by  ?/  square. 

6.  Write  the  square  root  of  the  sum  of  x  and  y  equals  c 
minus  the  square  root  of  the  sum  of  x  and  b. 

7.  Write  the  fraction,  the  binomial  3  times  x  plus  1,  di- 
vided by  5  times  x,  minus  the  fraction  3  times  the  bi- 


NOTATION.  25 

nomial  x  minus  1,  divided  by  the  binomial  Zx  plus  2,  is 
greater  than  9  divided  by  \\x. 

8.  Write  the  square  root  of  the  fraction,  h  divided  by  a  plus 
X,  plus  the  square  root  of  the  fraction  c  divided  by  a 
minus  x^  equals  the  4th  root  of  the  fraction,  4  times  the 
product  of  6  and  c,  divided  by  a  square  minus  x  square. 

9.  Write  a  exponent  f ,  minus  h  exponent  —  m.  Write  the 
result  in  three  different  forms. 

Til 

10.  Write,  a  exponent  -,  is  to  the  binomial  6  minus  x  ex- 

n 

ponent  —  f ,  as  5  times  c  square  plus  d,  is  to  the  5th 
root  of  X  4th  power. 

™  1  __ 

Result,  a»  :  -   :  :  ^C^  -\-  d  :  ^j;4 

What  binomials  are  there  in  the  last  result  ? 


EXERCISES    IN    READING   AND    EVALUATING  EXPRES- 
SIONS. 

Eead  the  following  expressions  and  find  the  value  of  each  ; 
if  a  =  6,  6  =  5,  c  =  4,  and  d  =  1. 

1.  a^  +  2ab  —  c  +  d. 

•  BesuU,  36  +  60  —  4  +  1,  or  93. 

2.  2a3  _  Sa^ft  +  c\  BesuU,  —  44. 

3.  3(a2  —  62)  —a{c'^  4  d).  i.   idt,  15. 

4:0,0 

4.  Between  the  expression —^ and  s/^g^        (g-^   i    ±i)\ 

which  of  the  signs,  =,  ^,  or  <^  is  correct  ? 

Read  and  evaluate  the  following,  calling  a  =  16,  5  =  10, 
c  =1  G,  m  =  i,  X  =  5,  and  ?/  =  1. 


5.   (6  —  x){Va+  b)  +  ^{a—  b){x-{-y). 

ReaulK  76. 


26  FUNDAMENTAL    RULES. 

/ 

6.  Vc  {a  +  0)  —  \^c^  {a  —T).       Result,  6.49,  nearly. 

7.  50aa;-^  +  4.a^  —  100  [^  -  {^x  +  ^)]. 

Result,  —112. 

8.  3«-  i  +  4  /^^i:^"  +  5-5^  +  ^) .       Result,  41. 

Suggestion. — Any  power  of  1  is  1. 

9.  ^x  -  ^^~^'  +  f-  -  rt-'  Z»l  i?cs?^/if,  2451. 

X  —  y        b  - 


10.  Find  the  value  of  a^x'  —  da  -\-  xVx^  +  3a,  when 
X  z=  o,  and  «  =:  8. 

11.  Find  the  value  of  <^  +  dVix  +  «/)  —  {a—b)\/(x—y), 
when  a  =  10,  Z*  =  8,  i?;  =  12,  and  y  =  4. 

SuG.     \^(x  +  y)  is  equivalent  to  \^x  +  y,  the  parenthesis  and  the 
vinculum  having  the  same  signification. 

12.  If    a  =  2,    I?  =  3,    X  =  6,    and    y  =  6,    show    that 

=  9. 

13.  With  the  same  values  show  that  (ayy(bx-\-a^-\-d)~a 

{b{x-  v)-i  -  [{axY  -  142]  j  20^^  _ 
{b-ar 

14.  Find  the  vaUie  of  VlO  +  n  —  (10  +  7i)i  if  w  =  6. 

15.  Find  the   value   of    (5m^  +  5Vcc)^  +  Vm  +  x^,    it 
??z  rrr  4,  and  a;  =  9. 

Test  Questions.— What  are  the  chief  points  of  diflference  between 
the  Arabic  or  Decimal  notation,  and  the  Literal  or  Algebraic?  What 
is  meant  by  the  terms  positive  and  negative  as  applied  to  quantity  ? 
What  is  the  meaning  of  the  negative  sign  when  prefixed  to  an  expo- 
a^  —  b' 


nent  ?    Read  —         —  CX ,  or  =,  or  <,  or  >  a/ mx  —  y-'^  :  im^. 


NOTATION. 


27 


Synopsis   for   Eeview. 


What? 


o 

C 


o 


0/ 
(Quantity 


Arabic. — See  Ai-ithmetic. 

(1.  More  general. 
Advantages.  |  2.  Can  trace  quant's. 

Examples. 

{  Known  quantities  j  g^j^ 
^  I  Unknown       "         ( 

Literal  ^  ^'^  ^^"'^      Acc'ts,  subs's,  Gr.  let's. 
[Diff.  bet.  Alg.  andArith. 

["Letters    in  connection. 
2nd  Law  \  Sch.  1.  Two  p'ts  of  diff. 
[  Sell.  2.  Fig's  with  lett'rs. 

Infinity,  meaning  of. 


0/  Operation 


X,  4-,   :  ,  p   h)a{c,  a\h__ 
c 

Definition,  Power,  Root,  ^. 

\  Integral, 
Exponent,  \  Fractional,  !-  How  read. 
sarative. 


iNe: 
,  =,::,  X,  a. 


Of  Relation  |  :,  •• 
Of  Aggregation  \- 
Of  Continuation  \ , . 

Of  Deduction  |  .•.,'.•  . 

What  ?     ///.     How  distinguished. 

Two  signs  of  every  quantity. 

Plus  and  Minus  Terms  become  Positive  and  Negative 

C  Meaning. 
"Less  than  zero.  "J  How  negatives  increase. 

[  How  a  quantity  changes  sign. 

o  ^  r 

2      Polynomial.     Term, 
i  g  J  Monomial.     Binomial.     TrinomiaL 
o  g   j  Coefficient. 
^fg     Similar  Terms. 


28  FUNDAMENTAL   RULES, 

SECTION  IL 
Addition. 

S4:,  Addition  is  the  process  of  combining  several 
qaantities,  so  that  the  result  shall  express  the  aggregate 
value  in  the  fewest  terms  consistent  with  the  notation. 

6S.  The  Su7n  or  Amount  is  the  aggregate  value 
of  several  quantities,  expressed  in  the  fewest  terms  con- 
sistent with  the  notation. 

III. — To  add  346,  234,  and  15,  is  to  find  an  expression  for  their  ag- 
gregate value  in  the  fewest  terms  consistent  with  the  decimal  notation. 
The  sum  or  amount  is  595,  because  it  is  such  simplest  expression  for 
the  aggregate.  In  hke  manner  the  sum  of  4ac  -)-  5&  +  ^x,  13ac  -|-  26 
+  8x,  and  126  +  9^,  is  17ac  -\-  19x  -J-  196,  because  it  is  the  simplest 
expression  for  the  aggregate  value  consistent  with  the  literal  notation. 

The  aggregate  value  of  346,  234,  and  15,  is  expressed  by  simply  read- 
ing them  in  connection;  as,  "346  and  234  and  15."  But,  letting  h 
stand  for  hundi'eds,  t  for  tens  and  u  for  units,  this  is  (3/i,  +  4<  -j-  6?i) 
+  (2/i  -f  3i  ,-f  h.1)  -\-  {It  -f-  5m)  or  a  polynomial  having  2  trinomial 
and  one  binomial  terms,  or  8  terms  in  all.  But  595  is  ^h  -\-  ^t  -\-  5u, 
or  simply  a  trinomial. 

If  the  pupil  is  acquainted  with  other  scales  of  natation  he  knows 
that  with  radix  100,  595  is  expressed  by  2  figures. 


6G,  JProp.  !•  Similar  terms  are  united  by  Addition 
into  one. 

Dem. — Let  it  be  required  to  add  4ac,  5ac,  —  2ac,  and  —  3ac.  Now 
4ac  is  4  times  ac,  and  5ac  is  5  times  the  same  quantity  (ac).  But  4 
times  and  5  times  the  same  quantity  make  9  times  that  quantity. 
Hence,  4ac  added  to  5ac  make  9ac.  To  add  —  2ac  to  9«c  we  have  to 
consider  that  the  negative  quantity,  —  2ac,  is  so  opposed  in  its  char- 
acter to  the  positive,  9ac,  as  to  tend  to  destroy  it  when  combined 


ADDITION.  29 

(added)  with  it.  (As  if  9ac  were  property,  and  —  2ac  debts. )  There- 
fore, —  2ac  destro^-s  2  of  the  9  times  ac,  and  gives,  when  added  to  it, 
7ac.  In  hke  manner  —  Sac  added  to  lac,  gives  4ac.  Thns  the  four ' 
similar  terms,  4ac,  5ac,  —  2ac,  and  —  Sac,  have  been  combined 
(added)  into  one  term,  4r/c  ,•  and  it  is  e\-ident  that  any  other  group  of 
similar  terms  can  bo  treated  in  the  same  manner,     q.  e.  d. 

EXAMPLES. 

1.  Add  ISm^n,  —  lOm-n,  —  6771271,  Bm'^n,  and  —  Am^n. 

Model    Solution. — Adding   together  13m^n,   and  —   lOm^n,    the 

—  lOm^n  destroys  10  of  the  13  times  m'^n  and  gives  3m^n.  Add- 
ing 3m^7i  and  —  Qm^n,  the  3m^7i  destroys  3  of  the  —  Grn"^ ji  and 
gives  —  dm^n.  —  Sm^n  added  to  5m^n  destroys  3  of  the  5  times 
m^n  and  gives  2m^n.      2m'^n  added  to  —  Am^n  destroj's   2    of  the 

—  4m%  and  gives  —  2m'^n.     Hence,  the  sum  of  Idm'^n,  —  lOm^n, 

—  Gm^n,  5m^n,  and  —  4??i^n  is  —  27n'^n. 

2.  Add   18ax^,  —  5ax^,  — lOax^,  4:ax^  and  —  Gax'^,  ex- 
plaining as  above.  Result,  ax^^. 

3.  Add  —  5c ^^2^  —  26^ x"',  Sc^x^-,  3c^.r2,  and  —  4:C^x'^,  ex- 
plaining as  before.  Result,  0. 

4.  Add  3ajT,  ^ax,  —  ax,  2ax,  —  lax,  and  ^ax. 

5.  Add  2hy\  —  (Sbif,  —  hy\  8by\  3by\  and  —  2by^. 

6.  Add  5ax-^,  —  2ax^,  3ax-\  —  dax'^,  and  ax^. 

Sum,  —  1ax\ 

7.  Add  5;r^,  —  Qx^,  —  10^^,  +  3x^,  and  llA 

8.  Add  —  Ga2,  +  2a\  —  Sa^,  4.a\  —  Za\  and  a\ 

9.  Add  —  2a^x,  -f  a^^x,  —  3av^jr,  la^x,  and  — 4.a^x. 

■  Sum,  —  a^x. 

10.  Add  —  20771^,  4aV^,  3a77i^,  and —  a'^m. 

Sum,  4:a'y~m. 

11.  Add  lOa^x^,  ~4.a^Vx,—  2a^x^  and  4.a^v'x. 


30  FUNDAMENTAL   RULES. 

12.  Add  l^m,  l^am,  —  3a/n,  and  am.  Sum,  2am. 

13.  Add  21a-,  —  a\  —  28^2,  and  —  4a2.  Sum,  —Ga^ 

14.  Add  3w^  —  ^m\  m-,  and  —  ^m'^.  Sam,  2-^-^mK 

15.  AddTv^^,  —  5^x,  12V x,  and  ^  ^^x.  Sum,  11  v"^. 

16.  Add  96,  lb,  —  f  6,  —  86,  —  |6.  Sum,  —^b. 

G7*  Cor.  1. — In  adding  similar  termi^,  if  the  terms  are  all 
positive,  the  sum  is  positive  ;  if  all  negative,  the  sum  is  nega- 
tive; if  some  are  positive  and  some  negative,  the  sum  takes  the 
sign  of  that  kind  (positive  or  negative)  which  is  in  excess. 

ScH. — The  operation  of  adding  positive  and  negative  quantities 
may  look  to  the  pupil  like  Subtraction.  For  example,  we  say  -4-5 
and  -3  added  make  -t-2.  This  looks  like  Subtraction,  and,  in  one 
view,  it  is  Subtraction.  But  why  call  it  Addition  ?  The  reason  is, 
because  it  is  simply  putting  the  quantities  together — aggregating  them — 
not  finding  their  difference.  Thus,  if  one  boy  pulls  on  his  sleigh  5 
pounds  in  one  direction,  while  another  boy  pulls  3  pounds  in  the  op- 
posite direction,  the  combined  (added)  effect  is  2  pounds  in  the  direc- 
tion in  which  the  first  pulls.  If  we  call  the  direction  in  which  the  first 
pulls  positive,  and  the  opposite  direction  negative,  we  have  -+-5  and 
-3  to  add.  This  gives,  as  illustrated,  -t-2.  Hence  we  see,  that  the 
sum  of  -t-5  and  -3  is  -i-2. 

But  the  difference  between  +5  and  —3  is  8,  as  appears  in  the  fol- 
lowing illustration  :  Suppose  one  boy  is  drawing  his  sleigh  forward, 
while  anotlieris  holding  back  3  lbs.  If  it  takes  just  10  lbs.  to  move 
the  sleigh  itself,  the  first  boy  will  have  to  pull  13  lbs.  to  get  it  on.  But 
if  instead  of  holding  hack  3  lbs.,  the  second  boy  pushes  5  lbs.,  the  first 
boy  will  only  have  to  pull  5  lbs.  Thus  it  appears,  that  the  difference 
between  pushing  5  lbs.  (or  +5)  and  holding  back  3  lbs.  (—3)  is  8  lbs. 
In  like  manner  the  sum  of  $25  of  property  and  $15  of  debt,  that  is 
the  aggregate  value  when  they  are  combined,  is  $10.  -+-25  and  -15 
are  -t-10.  But  the  difference  between  having  $25  in  pocket,  and  being 
$15  in  debt,  is  $40.     The  difference  between  -+-25  and  -15  is  40. 

17.  A  thermometer  indicated  -+-28°  (28°  above  0);  it  then 
rose  10°,  then  fell  3°,  then  rose  2°,  and  again  feU  7°. 
What  was  the  sum  of  its  movements ;    or,  how  did  it 

stiind  at  last? 


ADDITION.  31 

Model  Solution. —  Calling  upward  movement  -t- and  downward-, 
the  movements  were  -t-10,  -3,  -t-2  and  -7,  the  sum  of  which  is  ^-2. 
Hence  it  rose  2°.  As  it  originally  stood  at  28°  above  0,  and,  in  the 
v/hole,  rose  2^,  it  stands  at  last  30^^  above  0. 

18.  A  party  are  rowing  up  a  stream,  and  alternately  row 
and  rest.  During  3  periods  of  rowing  they  advance 
Smn,  2mn,  and  6mn  rods.  But  during  the  correspond- 
ing periods  of  resting,  they  float  down  5mn,  mn,  and 
4m7i  rods.  What  was  the  result ;  did  they,  on  the 
whole,  ascend  or  descend,  and  how  much  ?  In  other 
words,  what  is  the  sum  of  -^^mn,  -h2mn,  -i-6mn, 
-5mn,  -mn,  and  -imn  ? 

Ans.  The}^  ascended  mn  rods.   (  -hmn.  ) 

19.  A  man  has  a  farm  worth  $100c^,  on  which  there  is  a 
mortgage  of  $lDcd ;  he  has  personal  property  worth 
$8cd,  and  accounts  due  him  of  $2cd,  but  owes  on  ac- 
count $5cd,  and  on  note  ^Icd.  What  is  the  sum  of  his 
effects?  Or  what  is  the  sum  of  -+100cd,  -15cd, 
■+8cd,  -t2cd,  -5cd,  and  -led  ? 

Ans.  He  is  worth  $83cd.   {-hSScd.) 

OS,  CoE.  2. —  The  sum  of  two  quantities,  the  one  positive 
and  the  other  riegative,  is  the  numerical  difference,  luith  the 
sign  of  the  greater  prefixed. 

60,  CoR.  3. — It  appears  that  addition  in  mathematics  does 
not  always  imply  increase.  Whether  a  quantity  is  increased 
or  diminished  by  adding  another  to  it  depends  upon  the  relative 
nature  of  the  two  quantities.  If  they  both  tend  to  the  same 
end,  the  result  is  an  increase  in  that  direction.  If  they  tend 
to  opposite  ends,  the  result  is  a  diminution  of  the  greater  by 
the  less. 


70,  Pvojy*  2,  Dissimilar  tei^ms  are  not  united  into  one 
by  addition,  but  the  operation  of  adding  is  expressed  by  lorit^ 


32  FUNDAMENTAL   RULES. 

ing  them  in  succession  with  the  positive  terms  preceded  by  the 
+  sign  and  the  negative  by  the  —  sig7i. 

Dem. — Let  it  be  required  to  add  -|-  ^cy^,  +  3a&,  —  2xy,  and  —  mn. 
4:cy^  is  4  times  cy^,  and  3ab  is  3  times  ab,  a  different  quantity  from 
cy^  ;  the  sum  will,  therefore,  not  be  7  times,  nor,  so  far  as  we  can  tell, 
any  number  of  times,  cy^  or  ab,  or  any  other  quantity,  and  we  can 
]onlj  express  the  addition  thus  :  4:cy^  -\-  Sab.  In  Uke  manner,  to  add 
rto  this  sum  —  2xy  we  can  only  express  the  addition,  as  4c?/ ^  -}"  3a&  + 
( —  2xy).  But  since  2xy  is  negative,  it  tends  to  destroy  the  positive 
quantities  and  will  take  out  of  them  2xy.  Hence  the  result  will  be 
4ct/^  -{-  Sab  —  2xy.  The  effect  of  —  imi  will  be  the  same  in  kind  as 
that  of  —  2xy,  and  hence  the  total  sum  will  be  4c?/^  -|-  Sab  —  2xy  — 
inn.  As  a  similar  course  of  reasoning  can  be  appUed  to  any  case,  the 
truth  of  the  proposition  appears,     q.  e.  d. 

ScH. — In  such  an  expression  as  4ci/^  -|-  Sab  —  2xy —  mn,  the  —  sign 
before  the  7nn  does  not  signify  that  it  is  to  be  taken  from  the  imme- 
diately preceding  quantity  ;  nor  is  this  the  signification  of  any  of  the 
signs.  But  the  quantities  having  the  —  sign  are  considered  as  opera- 
ting to  destroy  any  which  may  have  the  -\-  sign,  and  vice  versa. 

EXAMPLES. 

1.  Add  together  5ax,  —  lOcy,  8b,  and  —  n. 

Model  Solution.  5ax  and  lOcy  being  dissimilar  will  not  unite 
into  one  term,  since  one  is  5  times  a  certain  quantity,  and  the  other  is 
10  times  cy,  a  different  quantity  ;  therefore  I  can  only  express  the  ad- 
dition, as  5aic -|-  (  —  lOcy.)  But  the  lOcy  being  negative  tends  to 
destroy  positive  quantities,  and  will  take  out  of  such  its  numerical 
value.  Hence  5ax  -{-  ( —  10c?/)  is  5ax  —  lOcy.  To  this  adding  86 
which  is  positive  and  hence  will  go  to  increase  the  result,  I  have  5ax 
—  lOcy  -\-  86.  Finally,  as  n  is  negative  it  diminishes  the  result  by  its 
numerical  value,  and  I  have  for  the  sum  5ax  —  lOcy  -\- 8b  —  n. 

2.  Add  together  4am,  —  2c'^y,  —  8x,  and  5bn,  explaining 
as  above. 

'  3.  Add  —  2c2?w*  +  4c??i,  —  Gc^m^,  and  lOc^m,  explaining 
as  before. 


ADDITION.  33 

4.  Add  7a  -^b,  —  S2x  '^,  Gmn,  and  —  5a^,  explaining  as 
before,  and  find  the  numerical  value  of  the  result  if 
a  =  3,  b  =  18,  X  =  8,  m  =  2,  and  n  =  b. 

Numerical  result,  21. 

71»  Cor. — Adding  a  negative  quantity  is  the  same  as  sub- 
r acting  a  numerically  equal  positive  quantity ;  that  is,  m 
+  (-nj  is  m  —  n,  shown  as  above. 


72,  JPi^ob, — To  add  polynomials. 

RULE. — Wkite   the    polynomials   so  that  similar 

TERMS  SHALL  FALL  IN  THE  SAME  COLUMN.  COMBINE  EACH 
GROUP  OF  SIMILAR  TERMS  INTO  ONE  TERM,  AND  WRITE  THE 
RESULT  UNDERNEATH  WITH  ITS  OWN  SIGN.  ThE  POLYNO- 
MIAL THUS  FOUND  IS  THE  SUM  SOUGHT. 

Dem, — As  the  object  is  to  combine  the  quantities  into  the  fewest 
terms,  it  is  a  matter  of  convenience  to  write  similar  terms  in  the  same 
column,  as  such,  and  only  such,  can  be  united  into  one.  {06^  70,) 
Now,  since  in  poljTiomials  the  plus  and  minus  terms  stand  in  the 
same  relation  to  each  other  as  positive  and  negative  quantities  {57), 
they  may  be  considered  as  such.  The  partial  sums  will  then  be  dis- 
similar terms  and  -wall  be  added  by  Prop,  Art,  70;  that  is,  by 
connecting  them  with  their  own  signs,     q.  e.  r>. 

EXAMPLES. 

1.  Add  16ac  —  2m  +  xy,  8m  —  bxy  —  d  —  2ac,  —  3xy 

—  4ac  —  Q>my  and  2mn  —  3ac  +  8xy. 
Model  Solution. — Writing  the  first  polynomial  as  it  stands,  I  ar- 

16ac  —  2m     +  xy  ^^^g^  *^^  «t^e^«  ««  t^^* 

—  2ac  +  3m  —  5xy  —  d  ^^^^^^  t^^^  ^^""^  ^^^^^ 
^^^ g^^^ 2j,y                                 in  the   same   column,    for 

_  ^ac  +  8^?/  +  2mn  convenience      in     uniting 

= ; ^^ 1 — ] — -; them.      There    beinsr     no 

lac,  —  5m,  +  xy,  —  rt,  +  2mn  ^  •    -i      *      i    o       t 

' 1 — \ iii —- term  similar  to  +  2mn  I 

lac   —  5m    +  xy  —d   +  2mn      Turing  it  down,  and  m  hke 
manner  —  d.     8xy  and  —  Sxy  are  5xy.     Bxy  and  —  ^xy  are  0.     0  and 


34  FUNDAMENTAL   liULES. 

xy,  or  simply  -\-  xy,  is  the  sum  of  the  similar  terms  in  xy.  Writing 
this  result  I  pass  to  the  next  column.     —  6m  and  -J-  3m  are  —  3m. 

—  3m  and  —  2m  are  —  5m  which  being  the  sum  of  the  similar 
terms  in  m,  is  written  down.  In  hke  manner  the  sum  of  the  terms 
in  ac  is  lac.     The  partial  sums   are,  therefore,    lac,  —  5m,   -|-  xy, 

—  d,  and  -\-  2mn.  But  these  being  dissimilar  terms  are  added  by 
connecting  them  with  their  own  signs,  (70);  whence,  the  sum  of 
the  several  polynomials  is  lac  —  5m  -{-  xy  —  cZ  -f  '^mn. 

In  like  manner  solve  and  explain  the  following  : 

2.  Add  6x  +  5ay,  —  3x  -\-  2ay,  x — •  6a?/,  and  2j7  +  ay. 

Sum,  ^x  4"  2ai/. 

3.  Add  3a?/  —  7,  —  ay  +  8,  lay  —  9,  —  3a?/  — 11,  and  lOay 

—  13.  •  Sum,  Hay  — 32. 

4.  Add  — 3a6  +7x,  3a6  — 10^,  3a6  —  Q>x,  —ah  +  ^x,  and  2a6 
+  4^.  Sum,  Aab  +  4r. 

5.  Add  —  6a''  +  26,  —  35  +  2a^  —  5a^  —  Sb,  Aa'~  —  2b,  and 
db  —  3aK  Sum,  —  Sa^— 26. 

6.  Add    3a"-b^  —  706^  +  ^cixy,  —  la'^b^  —  2ab'  —  axy,    ab* 

—  laxy  +  8a263,  —  10a6^  +  a-6-^  +  3axy,    and  —  Sa'^b^ 
+  18a6<.  Sum,  0. 

7.  Add  13a^2  —  14^2  4.  ^ac^  —  mn'-,  4.ax^  +  Wy^  —  da^c, 
4:y^  —  17a^2  -j.  2ac^  +  2m'^n  +  3mn-,  and  lOy^  —  a^c. 

Sum,15y'  +  5ac^  +  2mn^  —  4a3c  +  2m^n. 

8.  Find  the  sum  of  2a3  -|-  4:¥x  —  c'x%  2c'-x''  +  4a^  —  66% 
and  26^07  —  Ac^x^-  -\-  2a\  Sum,  8a^  —  3c"-x^. 

9.  Find  the  sum  of  Sa'^x^  —  Sxy,  5ax  —  5xy,  9xy  —  5ax, 
2a"x--\-xy,  and  5ax — 3xy.     Sum,  lOa^a;^  -\-  5ax  —  xy. 

10.  Find  the  sum  of  2bx—  12,  3^^— 26j:,  5x^—3^''^   6^x 
+  12,  x-^+3,  and  5x-^  —  l\/x. 

Sum,  14a;2  —  4:^x-\-S. 

11.  What  is  the  sum  of  20a^c^x  -f  15a/i  —  15  a-'c^x  —  23a/i  ? 

Ans.,  ba'^c^x  —  8a/i. 


ADDITION.  35 

12.  What  is  the  sum  of  16xy  —  4/im  +  llxy  +  43/i?n  ? 

Ans.,  3dxy  +  ddhm. 

13.  What  is  the  sum  of  Bc^  —  49an  +  lOc^  +  l^a/i  ? 

^ns.,  15c*  —  35an. 

14  What  is  the  sum  of  lOx^y^  —  llsk  +  15^^?/*  +  5sk 

—  4tx'^y^  ?  ^ns.,  21x'^y'^  —  I2sh 

ScH.  1. — In  practice,  the  partial  sums  are  written  at  once  in  connec- 
tion, with  their  own  signs.  Also  it  is  desirable  to  avoid  repeating  the 
quantities  added.  And,  finally,  the  expert  will  not  take  the  trouble  to 
arrange  the  poljniomials,  but  will  simply  select  and  combine  the 
similar  terms,  wiiting  each  result  in  the  total  sum,  at  once.  Thus,  in 
solving  Ex.  7,  when  the  object  is  simply  to  find  the  sum,  and  not,  as 
above,  to  explain  the  process,  we  proceed  as  follows  :  Noting  the  13ax^, 
we  cast  the  eye  along  till  we  find  the  similar  term,  -|-  ^aic^,  and  say 
"  -|-  llax^  ;"  again,  casting  the  eye  along  till  we  find  —  Ylax^,  we  say 
"0."  Therefore  nothing  is  written  in  the  sum  for  these  tenns,  as  they 
mutually  destroy  each  other.  Again,  looking  to  —  14?/^,  and  then  on 
to  IS^/^,  we  say  "y^,"  and,  passing  on  to  -)-  ^y^,  say  "  5y2,"  and  again 
pass  on  to  \Qy^  and  say  "ISt/^."  This  being  the  sum  of  the  terms  in 
2/2,  it  is  then  written  in  the  answer.  In  the  same  manner  the  work  is 
carried  on  to  completion  :  i.  e. ,  ordy  naming  results,  and  writing  them 
in  the  total  sum. 

In  this  manner,  write  the  answers  in  the  following  ex- 
amples : 

15.  Add  3^  —  5?/  4-  4c,  2x  —  2y—  3c,  and  —  x  +  Sy  -{-  c. 

Sum,  4:X  —  4?/  +  2c. 

16.  Add  11a  +  ISx  —  Id,  4.a  —  lOx  —  2d,  and  —  9a  —  ^ 
+  3d  Sum,  6a  +  207  —  M. 

17.  Add  3a^  —  4thy  +  2mn  —  16,  2hy  —  5mn  -f  11,  3m?j 
— 2ax  +  5,  and  —  hy  -\-  mn  —  ax. 

Sum,  mn  —  Zby. 

18.  Add   6am^  —  36  +  ^cxy  —  2ax^,  Ab  —  Sexy   +  Soa:*, 
llcxy  —  xy  —  8amx  —  Sax'^,  and  5xy  —  6  —  b. 

Sum,  12cxy  —  2a772^  -f  Axy  —  6. 


36  FUNDAMENTAL    RULES. 

19.  Add  1x\j  —  2.i7-,  Zx^  +  xy,  x/^  +  xy,  and  4.r*  —  ^yx, 

20.  Add  2a^  —  30,  3^-^  —  2ax,  ^xr-  —  3^^,  and  Z^lc  +  10. 

/S'l^m,  8^:2  _  20. 

21.  Add  8a2j72  —  ^ax,  lax  —  ^xy,  9xy  —  5ax,  and  2a^x^ 
+xy 

m.  Add  6a.r2  +  Wx,  —  2ax^  —  6x^,  Sax^-  —  IOj?^  —  7a^- 
H-  3v/  ^,  and  a^-2  +  llv"  x. 

23.  Add  6xy  —  12^^^  —  4^2  _^  3^^^  4^2  —  2jti/,  and  —  Sxy 

+  4^2. 

24.  Add  4.ax  —  130  +  3^^,  Bx^  +  3aa;  +  9^2,  7^?/  —  4v/^ 
-f  90,  and  v^^  +  40  —  6x^ 

25.  Add  3a-2  +  46c  —  r^  +  10,  —  5a-^  +  .66c  +  2e2_-15, 
and  —  4a- 2  —  96c  —  lOc"-  +  21. 

26.  Add  7a  — S?/'',  8v^'  +  2a,  5if  —  ^x,  and  —  9a  +  7v^ 
together.  Sum,  lA>/x. 

27.  Add  4m7i  +  3a6  —  4c,  3a;  —  4a6  +  2mn,  and  3m2  —  4p 
together. 

>S'a??i,  6m?i  —  a6  —  4c  +  3.r  +  3m2  —  4p. 

28.  Find  the  sum  of  3a2  +  2a6  +  46^  5a2  —  8a6  +  6^  —  a* 
+  5a6  —  62,  18a2  —  20a6  —  1962,  and  14a2  —  3a6 
+  2O62. 

29.  Find  the  sum  of  4:X^  —  ha^  —  5a^2  _|.  6a2^,  %a^  -f  3^^ 
+  4ajT2     -]-  2a^x,     —  Vlx^     +  19a^2    —  IBa^x,     I'^ax^ 

—  27a2^  +  18a^  3a2j7  —  20a'^  +  12.r3,  and  31a2.r  —  2^?^ 

—  31a^2  —  7^3.  Sam,  —  7^^  — a^. 

30.  Find  the  sum  of  2a6  +  12  —  xHj,  x^y  -\- xy  +  10,  Zxy^ 

+  2^2y  —  xy,  5xy  +  11  +  x^y,  and  17  —  2x^y —  x'^y 
Sum,  2ab  +  50  +  5xy  —  x'^y  -\-  4:xv'y. 


ADDITION.  37 

31.  Add  j-^  -[-  ojc  —  ah,ab  —  \^x  +  xy,  ax  -\-  xy  —  4a/j,  x'^ 
+  \/j;  —  X,  and  xy  -{■  xy  -{-  ax. 

Sum,  2x'^  -\-  Sax  —  4a6  +  ^^y  —  ■^• 

32.  Add  Ix^y  —  2xv'^  +  7,  ^xy  +  3xy^  +  2,  SyV^-—  v^^ 

,-1  1.  A 

—  6,  9yva;  —  4i/^j;  —  3,  and  1  +  Ixy'^  —  2yx^. 

Sum,  ISx^y  -f  Sxy^  +  1. 

ScH.  2. — The  object  and  process  of  addition,  as  now  explained, 
will  be  seen  to  be  identical  with  the  same  as  the  pupil  has  learned 
them  in  Arithmetic,  except  what  grows  out  of  the  notation,  and  the 
consideration  of  positive  and  negative  quantities.  For  example,  in 
the  decimal  notation  let  it  be  required  to  add  218,  10506,  5003,  81, 
and  106.  The  units  in  the  several  numbers  are  similar  terms,  and 
hence  are  combined  into  one  :  so  also  of  the  tens,  and  of  the  hundreds. 
To  make  this  still  more  evident,  let  ii  stand  for  units,  t  for  tens,  h  for 
hundreds,  th  for  thousands,  and  t.  th  for  ten  thousands.  21'"  is  then 
2/i  +  1«  +  8m,  10506  is  Itth  +  5h  +  6u,  5003  is  5th  +  3u,  81  is  8«, 
+  lu,  and  106  is  Ih  -[-  6u.  "Writing  these  so  that  similar  terms  shall 
fall  in  the  same  column,  we  have  the  arrangement  in  the  margin. 
Whence,  adding,  we  get  the  sum.  The  process  of  carrying  has  no 
analogy  in  the  literal  notation,  since  the  relative  values  of  the  terms 

are  not  supposed  to  be  known. 
2h  -{-  4:t  -{-  oil         Again,  there  is  nothing  usually 
It.th  -{-oh  -\-  6u         found   in  the  decimal   addition 

^*h  -\-  oil         ii]£e  positive  and  negative  quan- 

of  -\-.lu         titles.     With  these   two   excep- 
tions the   processes   are   essen- 


l/l  4-  Qu 


It.th  +  Wi  +  9A  +  4/!  +  4?^         *'^^^^  ^^^  '^°'^-     The  same  may 

be  said  of  addition  of  compound 


1oJ44  numbers. 


73.  I^roj),  3,  Literal  terms,  which  are  similar  only  with 
reqject  to  part  of  their  factors,  may  be  united  into  one  term 
with  a  polynomial  coefficient. 

D^^i-  — Let  it  be  required  to  add  5aar,  —  2cx,  and  2mx.  These  terms 
are  similar,  only  with  respect  to  x,  and  we  may  say  5a  times  x  and 


38  FUNDAMENTAL  EULES. 

—  2c  times  x  make  (5a  —  2c)  times  x,  or  (5a  —  2c)x.     And  then,  5a 

—  2c  times  a;  and  2m  times  x  make   (5a  —  2c  -|-  2m)   times  x,   or 
(5a  —  2c  -f-  2?n)x.     q.  e.  d. 

EXAMPLES. 

1.  Add  ax'\  —  bx^,  —  2cx'^,  and  4:mx-\  with  respect  to  xK 

Sum,  {a  —  b  —  2c  +  4??i)j72. 

2.  Add  4:xy,  Saxy,  —  lOmxy,  and  cxy,  with  respect  to  xy. 

Sum,  (4  +  3a  —  10m  +  c)j7j/. 

3.  Add  amx  +  2<i?/,  2c^  —  ddy,  and  3cZ^  +  5?/,  with  respect 
to  X  and  y.  Sum,  (am  +  2c  +  3c?)j7  +  (5  —  d)y. 

4.  Find   the  sum  of  ax'^    +  by^   -f  •^'^^^j   and   m^^   - —  ny^ 

—  pxy. 

Sam,  (a  -[-  m)x^  +  (^  —  '0?/'^  +  ('^  — 2^)^y- 

5.  Find  the  sum  of  aj;^  +  bx'^  +  c^-^?,  and  a^x^  —  b^x'^  —  c^x. 

Sum,  {a  +  a2)^3  ^  (^  —  53)^2  _j_  (^ —  c^)x. 

6.  Find  the  sum  of  (a  —  b  -\-  c)^ x,  (a  +  5  —  c)v^  x,  and 
(6  +  c  —  a)\^.  Sum,  (a  +  5  -f  c)  v"'  J-. 

7.  Find   the  sum   of   2ax-"'y^   +  36c  —  la  -  ^.r"^  +  36, 

3A^-'"  V'^  +  26c  +  -At  —  2^'  ^^^  -^    —  6   ^' 
2a-'^^       _  '^""' 

1 ^.  >S^t^?7i,  3(a  +  h)  -^-  H-  56c  +  f  •  a-2^~  3. 

3 

8.  Find  the  sum    of   2xy'  +  2x-"'y'^  —  2,/:3,  —  S^'jr^y^ 

+  2r'^a?-'"i/^  —  a  +  6^73,    3^4^^  _  26u--"*  ?/^   +  3a,  and 

—  2^3    ^cy. 


74,  I^rop,  4,  Compound  terms  which  have  a  common 
compound,  or  polynomial  factor  may  be  regarded  as  similar 
and  added  with  respect  to  that  factor. 

Dem,  5(a;2  —  y"^  ),  2(x2  —  y"-  )  and  —  'd{x'^  —  y"-  )  make,  when  added 
with  respect  to  {x^  —  y^  ),  4:{x"  —y-  ),  for  they  are  5  +  2  —  3,  or  4 
times  the  same  quantity  (x^  —  y"  ).  In  a  similar  manner  we  may 
reason  on  other  cases,     q.  e.  d. 


ADDITION.  39 


EXAMPLES. 


1.   Find   the    sum   of    ^x  +  n  —  S,r\    5x^  —   2^  x -{-  n, 
3v/a7  +  n  —  lx\  and  IQx^  +  Sv^j-  +  n. 

Sum,  lOv^j-  +  n  +  5xK 

%  Find  the  sum  of  6a  —  6(a  —  5)  +  7,  3a  +  12(a  —  6) 

—  8,  and  2(a  _  6)  —  3a  —  20. 

Sum,  6a  —  21  +  8(a  —  h). 

3.  Find  the  sum    of  7(m  -f  3)  —  16(77i  —  3),  8(7?i  +  3) 
+  7(m  —  3),  and  3(m  —  3)  —  4(m  +  3). 

>S'wm,  ll(m  +  3)  —  6(m  —  3). 

4.  Find  the  sum  of  i-Va  _  36  —  ^V^  __  36  +  iVa  _  36 

—  -J-^V'a  —  36.  ^'w?7i,  t\V«  —  36. 

5.  Add  3   (a  —   <i){x  +  r)  "^, 1'  '       '     — >  and 


5  (a  +  c)  (^  +  y?) 


.,         3r3a  +  c) 
Sum,  — ^- 


6.  Find  the  sum  of  a(a  +  6)  +  3v^a  —  x,  —  4a(a  +  6) 
+  7a(a  — ^)"^,  —  ^a^a  —  ^■+  lla(a  +  6,)  —  2a(a4-  6) 
—  2 (a  —  xy,  and  5a(a  +  6)  +  14v''a  —  x. 


7.  Add  a^ X — y  +  hxij  -\-  c{a  +  xy,  —  bxy  +  (a  +  c) 


{a-\-  x)"']-{x  —y)'^,  2hxy-\-{a — T)^x —  y  —  a{a-{-x)^ 
Sum,  2a{x  —  y)^  +  "Ihxy  +  2c(a  +  xy. 

8.  Show  that  ^x  -{-by  -{-ax  —  z  +  amy  +  c^x  +  dz  +  y 
=  (am +  6+  l)i/  +  (c  +  1)^^  +  {d  —  l)z  +  ao:. 


40 


FUNDAMENTAL   KULES. 


Defs, 


Props. 


Synopsis. 

r  Addition. 

Sum,  or  Amount. 

rCor.  1.  Sign  of  sum. 
I  Cor.  2.  Sum  of  Pos. 

1.  Similar  Terms.     Dem.   ■{      and  Neg. 

I  Cor.  3.  Addition  not 
[     always  increase. 
Sell.  Sign  of  term. 

2.  Dissimilar  Terms.  Dem.  \  Cor.    Add.  of    Neg. 

=  Sub.  of  Pos. 


Prob.  To  add  Polys.    Rule.     Dem. 


Schol.   1.     Practical 

method. 
Schol.   2.     Same  as 

in  Dec.  Nota. 


Props. 


f  3.  Terms  partially  similar. 
[  4t.  Compound  similar  terms. 


Test  Questions.  —  Does  addition  always  imply  an  increase? 
When  does  it  not  ?  When  does  it  ?  What  is  addition  ?  How  are 
similar  terms  added  ?  How  are  dissimilar  added  ?  Give  the  Bule 
for  adding  polynomials  and  demonstrate  it. 


SECTION  III. 
Subtraction. 

7^»  Subtraction  is,  primarily,  the  process  of  tak- 
ing a  less  quantity  from  a  greater.  In  an  enlarged 
sense,  it  comes  to  mean  taking  one  quantity  from  an- 
other irrespective  of  their  magnitudes.  It  also  compre- 
hends all  processes  of  finding  the  difference  between-  quan- 
tities. In  all  cases  the  result  is  to  be  expressed  in  the  few- 
est terms  consistent  with  the  notation  used.  The  quantity 
to  be  subtracted  is  called  the  Subtrahend,  and  that  fi'om 
which  it  is  to  be  taken  the  Minuend. 


SUBTRACTION.  41 

76,  Hie  Difference  between  two  quantities  is,  in 
its  primary  signification,  the  number  of  units  whicti  lie  be- 
tween them  ;  or,  it  is  what  mud  he  added  to  one  in  order  to 
produce  the  other.  When  it  is  required  to  take  one  quantity 
from  another,  the  difference  is  what  must  be  added  to  the  Sub- 
trahend in  order  to  produce  the  Minuend. 

ScH. — The  most  comprehensive  and  fundamental  notion  of  differ- 
ence is  this  :  HaAdng  reached  any  specified  point  in  a  scale  of 
numbers,  or  in  estimating  magnitude,  how  (in  what  direction),  and 
how  far  must  we  pass  to  reach  another  specified  point  in  the  scale  of 
numbers,  or  in  the  value  of  the  magnitude. 

III. — When  we  ask,  ""What  is  the  difference  between  3  and  8?"  we 
ordinarily  mean,  "How  far  (over  how  many  units)  must  we  pass  in 
reckoning  from  3  to  8  ?"  That  is,  * '  How  many  units  added  to  3  will 
make  8  ?" 

Let  us  use  the  following  device  to  illustrate  the  whole  subject. 
Consider  A  the  zero  point  on  the  line  BC.     Call  distances  to  the  right 


- 

"^ 

^ 

■    + 

B 

A 

1. 

^ 

c 

— m 

1       1 
,— 9,— 8,- 

1 
-7,- 

1       1       1        1 
-6,— 5,— 4,— 3,- 

-l 

-1, 

0 

1    1 

+1.+2 

,+3.+4,-t-5.+6 

1       1       1 
,+7,+8,-f9,+i 

1 

of  A  positive  (  -<-),  and  distances  to  the  left  negative  (-).  Also  call 
reckoning  towards  the  right  positive  and  towards  the  left  negative, 
from  any  point  on  the  scale. 

1st,  The  di'fference  between  3  and  8,  means  either,  how  far,  and  in 
what  direction  must  we  go  to  pass  from  3  to  8  or  to  pass  from  8  to  3  ? 
In  the  first  case  we  pass  5  to  the  right,  and  say  3  from  8  is  -t-5,  un- 
derstanding that  3  and  8  are  both  positive.  But  to  pass  from  8  to  3, 
we  pass  5  towards  the  left,  and  hence  say  8  from  3  gives  -  5. 
These  two  results   are  expressed  thus  :  8  —  3  =  5,  and  3  —  8  =:  -  5. 

2nd,  The  difference  between  -  3  and  -  8  means  either,  how  far,  and 
in  ichai  direction  must  we  pass  to  go  from  -  3  to  -  8,  or  from  -  8  to 
-  3  ?  To  pass  from  -  3  to  -  8,  we  pass  5  towards  the  left,  and  hence 
say  -  3  from  -  8  gives  -  5.  That  is  -  8  —  (-  3)  =  -  5.  But  to  pass 
from  -  8  to  -  3,  we  pass  5  to  the  right,  hence  -  3  —  (-  8)  =  -t-5. 

3d,  The  difference  between  -  3  and  -^8  means  either,  how  far  and 
in  what  direction  must  we   pass   to  go  from  -  3  to  -t-8,  or  from  -t-8  to 


42  FUNDAMENTAL   RULES. 

-  3  ?  In  the  first  case  we  get  -+-8  —  (-  3)  =  -t-ll,  since  we  pass  11 
to  the  right.  In  the  second  case  we  get  -  3  —  (  +8)  =  -  11,  since  to 
go  from  -f-  8  to  -  3  we  pass  11  to  the  left. 

In  each  and  all  of  these  cases,  the  question  is,  "What  must  be 
added  to  the  quantity  conceived  as  the  Subtrahend,  in  order  to  pro- 
duce the  Minuend?"  considering,  in  each  instance,  the  number  of 
units  between  the  two  given  terms  as  the  numerical  value  of  the  dif- 
ference, and  its  sign  as  determined  by  the  fact  as  to  whether  we 
reckon  to  the  right  (up  the  scale),  or  to  the  left  (down  the  scale),  in 
passing  from  the  Subtrahend  to  the  Minuend. 


77.  I^VOh,     To  perform  Subtraction. 

RULE. — Change  the  Signs  of  each  Term  in    the    Sub- 
trahend   FROM    +    TO   ,    OR   FROM     ■ —     TO     +,     OR     CONCEIVE 

THEM    TO  BE  CHANGED,  AND   ADD    THE    RESULT    TO    THE    MiNUEND. 

Dem.— Since  the  difference  sought  is  what  must  be  added  to  the 
subtrahend  to  produce  the  minuend,  we  may  consider  this  difference 
as  made  up  of  two  parts,  one  the  subtrahend  with  its  signs  changed, 
and  the  other  the  minuend.  When  the  sum  of  these  two  parts  is 
added  to  the  subtrahend,  it  is  evident  that  the  first  part  will  destroy 
the  subtrahend,  and  the  other  part,  or  minuend,  will  be  the  sum. 
Thus,  to  perform  the  expmple  :  . 

From  ^ax  —  G&  —  3(Z  —  4m 

Take  2«.t  -^  2b  —  M  +  8m  ]       jf   these 

Subtrahend  with  signs  changed  — 'lax  —  26  -j-  5(/  —  iim  j.  three  quan- 
Minuend,  bax  —  6/>  —  3(Z  —  4m  j  tities       are 

Difference,  3o.c  —  8f/ -j- 2(2  —  12m    added     to- 

gether, the  sum  will  evidently  be  the  minuend.  If,  therefore,  we  add 
the  second  and  third  of  them  (that  is  the  subtrahend,  with  its  signs 
changed,  and  the  minuend)  together,  the  sum  will  be  what  is  neces- 
sary to  be  added  to  the  subtrahend  to  produce  the  minuend,  and  henco 
is  the  difference  sought,     q.  e.  d. 

EXAMPLES. 

1.  From  4a?>  +  3c2  —  xy  subtract  ^xy  —  2z  +  2a6  —  c^. 
Model  Solution. — Writing  the  subtrahend  under  the  minuend  so 


SUBTRACTION.  43 

that  similar  terms  shall  fall  under  each  other,  for  the  convenience  of 
combination,  I  have 

Aah  +  3c2  —  xy 

2ah  -^    c^  +  3xy  —  2z 

2ab  -f  40'"'  —  ixy  -\-  'Iz 

As  it  is  immaterial  where  the  process  of  subtraction  is  commenced, 
since  there  is  to  be  no  "  borrowing,"  I  will  commence  at  the  right,  as 
in  the  decimal  notation,  for  analogy's  sake.  The  first  question  is, 
What  must  be  added  to  —  2^  to  produce  the  corresponding  term  in 
the  minuend?  This  is  evidently*^  2z,  as  there  is  no  corresponding 
term  in  the  minuend,  and  I  have  only  to  write  a  term  in  the  differ- 
ence which  Tvill  destroy  —  2z  when  added  to  it.  Passing  to  the  next 
term,  I  inquire.  What  must  be  added  to  +  3xy  to  produce  —  xy  ? 
First,  I  must  add  —  Sxy  (the  term  with  its  sign  changed)  in  order  to 
destroy  the  -{-  Sxy,  and  then  I  must  add  —  xy  in  order  to  make  the 

—  xy  of  the  minuend.     So  in  all  I  must  add  —  Sxy  and  —  xy,  or 

—  4ucy.  —  ixy  is,  therefore,  the  difference  sought.  Passing  to  the 
next  term,  What  must  be  added  to  —  c2  to  make  -\-  3c^?  I  must 
add  -|-  c^  (to  destroy  —  c^)  and  -f-  3c2  (to  make  up  the  required  term 
in  the  minuend),  or  in  all  +  c^  and  -|-  ^c^,  or  -|-  Ac'^  ;  which  is  the 
difference.  Finally,  the  term  to  be  added  to  2ab  in  order  to  make 
4a&  is  composed  of  the  two  parts  —  2ab  and  -f-  4a6,  which  make  2ab. 

It  thus  appears  that  2ah  -|-  4c^  —  4xy  -}-  2z  is  the  difference,  since 
it  is  what  must  he  added  to  the  subtrahend  to  produce  the  minuend. 

A  Shorter  Explanation. — The  difference  sought  may  be  consid- 
ered as  consisting  of  two  parts,  1st,  the  subtrahend  with  its  signs 
changed,  and  2nd,  the  minuend  itself.  The  sum  of  these  two  parts  is 
the  difference,  since  it  is  what  is  necessary  to  be  added  to  the  subtra- 
hend to  produce  the  minuend.  Therefore  conceiving  the  signs  of  the 
subtrahend  to  be  changed,  and  adding,  I  have  2a&  -j-  40^  —  'key  -\-  2z, 
as  the  difference  sought. 

[Note.— Let  the  pupil  solve  and  explain  the  following  examples, 
giving  the  reason  for  changing  the  signs  of  the  subtrahend  (to  get  a 
quantity  which  added  to  it  will  destroy  it)  and  the  reason  for  adding 
this  result  to  the  minuend.  Keep  clearly  in  view  the  facts  that  the 
difference  is  what  is  required  to  be  added  to  the  subtrahend  to  pro» 
duce  the  minuend,  and  that  this  difference  is  made  up  of  two  parts.] 

2.  From  4:X  —  3¥  +  2ry  —  11  take  20-  +  b^  -\-  2xij  —  5. 

Bern,,  2x  —  46"^  -  ^  6. 


44  FUNDAMENTAL   RULES. 

3.  From  15ax  -j-  2¥y  —  Ga^x'^  take  —  5ax  —  4:t2y  —  Sa^xK 

Bern.,  20ax  +  6b^ij  —  Sa-'xK 

4.  From  Sa-'b--'  —  3xy  +   15  —  2^xy  take  8  +  lOxrj 
—  Sah-^  +  2^xy. 

Bern.,   lla^6-2  —  ISxy  +  7  —  4:\^lry. 

5.  From  4:a^x^  —  3c  +  Ub^y'^  take  Sh'^y'^  +  4^2^^  —  2r/. 

.    Bern.,  4.b^y^  —  3c  +  2rf. 

6.  From  a^  —  2a6  +  b^  take  a^  ^  2ab  +  62. 

i?em.,  —  4a6. 

7.  From  a^  —  6^  take  o^  —  2a6  —  b\  Bern.,  lab. 

8.  From   24:xy^   —  14m?/   +  l^x'^y^  —  14  +   21xz^  tako 
Vlxy^  —  10m?/  —  4^^^?/2  +  20^2^  —  8. 

9.  From  llpmx^  —  18n^  +  19m^  —  24  take  7pm^2  —  47^3 

_|-  107714  —  17. 

10.  From  a^  +  2ab  +  b^  take  a^  —  2ab  +  62. 

11.  From  a^  +  3a-^6  +  3a62  ^  53  take  a^  —  Sa'^b  +  3a52  —  63. 

2  1.13.  2.  1.1  2 

12.  From  x'^  +  2,r^?/^  +  y'^  take  o:^  —  2x^y'^  +  2/"^. 


13.  From  6v^  +  2/  +  4ar^a;2take  3(ir  +  ?/)^  —  4:a^xK 

Diff.,  3v'^'  +  y  4-  8a2^«. 

Suggestion. — Eegard  "/x  -f  2/  as  one  qiiantity,  and  obserye  thai 
3(a;  -\-  y)-  IB  the  same  as  Sv'aj  +  y. 

14.  From   6a;  —  Bv^^  +  17 (a  +   &)    take   2x  +  7v/^ 
+  4(a  +  6.)  i>i^.,  4a;  —  lOV'l^j  +  13(a  +  b). 

15.  From  10  —  a  +  6  —  c+o;  take  a -\-  b  —  c  —  x  —  90. 

16.  From  Q>7n  -\-  ^n  —  §a  —  1  take  %n  —  ^/z  +  ^  —  2. 

Diff.,  4m  +  n  —  a  +  1. 


SUBTRACTION.  45 

ScH.  1. — When  several  poljTiomials  are  to  be  combined,  some  by 
addition  and  some  by  subtraction,  it  ^^ill  be  found  expedient  to  write 
them  so  that  similar  terms  will  fall  under  each  other,  wTiting  the 
several  subtrahends  with  their  signs  changed,  and  then  add  the  quan- 
tities as  they  stand. 

17.  From  the  sum  of   ^aif  —  26  +  ^ax,  2m/^  -\-  b  —  ax,  and 
36  —  a?/«  +  3  —  y,  subtract  Say^  +  46  —  ax  -]-  y. 

Result,  —  26  +  3aa7  +  3  —2y. 

18.  From  the  sum  of  3a-  +  xy'-  —  26?/,  5a-  +  %r,y^-  —  36?/, 
and  Zxy^  +  4a2  +  by,  subtract  2a^  —  xy'^  —  56?/  +  5. 

BesuU,  10a2  +  Sxy^  -^  by  —  5. 

19.  From  the  sum  of  a'^¥  —  3]/^  +  ^xy,  Za"¥  +  3?/^  —  2j7?/, 
and  by-  -f  3a'62  —  xy  -\-  Q,  subtract  0-6^  -{-  xy  —  y^  —  3. 

Result,  Ga%^  +  Gif  +  j??/  +  9. 

20.  To  5ax'^  —  7cb  +  8»i^  —  2c-",  add  3c6  —  4c-"  +  2a^2, 
then  subtract  lOwi^  —  5xy  +  3c6  —  12c  ~",  add  Gm^ 

—  Ilc6  +  3n  —  4:ax%  subtract  —  16aa7^  —  2m'^  +    4c6, 
add  5m  ^  +  Gc" ",  and  subtract  4:xy  —  2?i. 

Result,  Idax-'  —  22c6  +  ll??i^  +  12c-"  +  xy  +  5n. 

21.  To  ley  - 1  +  8ax  —  56,  add  46  —  2ey  "  f  +  ?n,  then  sub- 
tract Bax  —  4??i  +  3  and  —  Sax  +  5c y~  ^  —  6,  add  lOao? 

_   2. 

—  26  +  8?7i  —  3,  and  subtract  Sm  —  lOcy     ^  —  2m. 

Result,  lOcy  - 1  +  16aa7  —  36  +  12m. 

ScH.  2. — The  proficient  solves  such  examples  as  the  above,  and,  in 
fact,  all  kindred  ones,  without  re-writing  the  quantities.  Thus,  to 
obtain  the  result  in  Hr.  20,  he  looks  at  5f?x^,  casts  liis  eye  along  till  he 
sees  2ax^,  says  Tax*  ;  then  looks  forward  to  —  4:ax^,  and  says  3ax^ ; 
then  notices  —  16«.r^  in  a  subtrahend,  and  hence  thinks  of  U  as 
-\~l&ax^,  and  says  19ax^.  Again  taking  tip  —  7ch,  he  looks  along 
noticing  the  similar  terms  and  saj-s,  mentally,  —  4:cb,  —  7cb,  —  18c6, 
—  22c6.  Again,  he  takes  up  8?n^ ,  and  running  through  with  the  sim- 
ilar terms,  says,  —  2m^,  4?7i-,  6ni'^,  ll?7i^.  And  so  for  all  the  other 
terms. 


46  FUNDAMENTAL   RULES. 

This  is  the  practical  way  of  doing  such  examples ;  a^id  the  pupil  should 
exercise  himself  till  he  can  write  out  the  result  at  once.  He  cau  gc  o\er 
the  preceding  examples  thus,  and  should  also  E-olve,  without  writing, 
those  which  follow. 

22.  From  13x-'  —  lax  —  952  take  5^^  —  ^ax  —  h\ 

23.  From  20a.^  —  v'^  -f  M  take  ^ax  +  5.r^  —  d. 

24.  From  5«6  +  262  _  c  +  5c  —  6  take  5-^  —  lab  +  he. 

25.  From  ax  -'^  —  ax  -'^  A;-  ex  —  d  take  ax:^  —  ax  -^  —  ex 
■—  M.  Biff.,  a{x  -3  —  ^3)  +  (c  4-  e)x  +  d. 

Suggestion. — This  would  at  first  become  ax -^  —  ax^  -\-  ex  -\-  ex 
-j-  d.     But  this  may  evidently  be  written  as  above. 

26.  From  x'^y  —  3\^xy  — •  6ay  take  SxH/  +  ^{xy)'-^   —  4ay- 

27.  From  the  sum  of  4a.r  —  150  +  4.r^  5x-^-i-  Sax  +  10  v^^, 
and  90  —  2ax  —  12  v^^  ;  take  the  sum  of  2ax  —  80 
+  lx%  Ix^  —  Sax  —  70,  and  30  —  4^^  —  2xi 
4-  4a2^2.  Result,  Wax  +60  —  x'^  —  ^a'^x'^. 

7S.  CoE.  1. —  WJien  a  par^enthesis,  or  any  symbol  of  like 
signification  (SO),  occurs  in  a  polynomial,  preceded  by  a  — 
sign,  and  the  parenthesis  or  equivalent  symbol  is  removed,  the 
signs  of  all  the  terms  which  ivere  within  must  be  changed,  since 
the  sign  —  indicates  that  the  quantity  within  the  parenthesis  is 
a  subtrahend. 

28.  Remove  the  parenthesis  fi'om  the  polynomial  3a^  x  -f 

252f/2  —  {^a^x  —  2m22;  +  Sb-y-)  and  express  the  result 
in  its  simplest  form.  Do  the  work  mentally,  writing 
only  the  result. 

Besult,  2m'^z  —  2a^x  —  GbH/K 

29.  Remove  the  parenthesis  from  5a  —  45  +  3c  —  (  —  3a 
4-  25  —  c).  Result,  8a  —  65  +  4c. 


SUBTRACTION.  47 

30.  Remove  the  parenthesis  from  4a  —  5x  —  (a  —  4^;)  -\- 
[x  —  8a).  Result,  —  5a. 

Queries. — In  Kt.  28  is  the  sign  of  Sa^x  changed  ?  What  is  its  sign  as 
it  stands  in  the  parenthesis  ?     Is  the  —  sign  which  appears  in  the 

result  before  5a^a',  the  same  as  the  one  before  the  parenthesis  in  the 
example  ?  No.  What  became  of  that  before  the  parenthesis  ?  Ans. 
The  operation  which  it  indicated  having  been  performed,  it  is  dropped. 
In  Ex.  30,  why  are  not  the  signs  of  the  terms  in  the  last  parenthesis 
changed  ? 

to*  Cor.  2. — Amj  quantity  can  he  placed  within  a  paren- 
thesis, p)receded  by  the  —  sign,  by  changing  all  the  signs.  The 
reason  of  this  is  evident,  since  by  removing  the  parenthesis 
according  to  the  preceding  corollary,  the  expression  would 
return  to  its  original  form. 

31.  Introduce  within  a  parenthesis  the  3d,  4th  and  5th 
terms  of  the  following  expression  :  Q>ax  —  ^cd  —  8??j  f 
5^  —  1y  -\-  X  —  4a. 

Besult,  Gax  —  2cd  —  (8m  —  5x  -\-  2y)  -}-  x  —  4a. 

32.  Introduce  within  a  parenthesis  the  last  three  terms  of 
^xy  -\-2cb  —  8x  —  5  +  2b. 

Result,  Axy  +  2cb  —  {8x -\-  5  —  2b). 

33.  Include  in  brackets  the  3d,  4th,  and  5th  terms  of  hax 
—  2.r2  4-  3a7  —  12ay  +  15.  Also  the  4th  and  5th. 
Also  the  2d  and  3d. 

Form  of  the  first  and  last,  5ax  —  2x'^  +  (3^7  —  12ay  + 
15),     5ax  —  {2x^~  —  ^x)  —12  ay  -\-  15. 

Query. — In  the  last  is  the  sign  of  2x*  changed  ? 

34.  Prove  that  (3^  —  &y)  +  {4.y  _  4^)  +  2  (^  +  2y)  =. 
x-^2y. 

35.  Prove  that  ^  (a  +  6  —  c)  +  |(6  +  c  —  a)  =  &. 

36.  Prove  that  6a  —  46  —  2  (a  +  Z>)  =  2  (2a  —  36). 


48  FUNDAMENTAL    RULES. 

37.  Prove  that  x^  +  6x^y  +  3  —  (^^  +  4:X^y    +    1)  = 

2{x^j  +  1). 
*38.  Prove   that  ^(a  —  56  +  ^c)  +  i(56  —  2a  +  ^c)  = 

ic  —  -^^b  —  ^a. 

39.  Prove   that   ^(a^  _  «   +  i)   4.  i(2a2  _  ^  _|_  2)  = 
-iL(10a2_7a  +  10). 

40.  Prove  that  ^a  +  ^b  —  {^a  —  ^b)  =  b, 

41.  Prove  that  ^(9  —  15^)  —-^(12  —  20^)  —  1(45^;  —20) 
=  1  —  4^. 

42.  Prove   that   ^{a  —  ^b  +  ^c)   +  ^{a  —  ^b  +  ^c)  == 
^_(90a  _  366  +  25c). 

SO.  CoR.  3, —  When  several  parentheses  occur,  included  the 
one  within  the  other,  begin  the  reynoval  ivith  the  inside  one. 

43.  Eemove  the  parentheses  and  other  marks  of  aggrega- 
tion from  4a  —  { —  [c  — ■  d  -{-  (4^^  —  i)  —  j^y-j  — 3iy|. 

Result,  Aa  -{-  c  —  d  -{-  4^2  —  1  —  xy  -\-  dy. 


44.  Show  that  a  —  [b  —  {c  —  {d  ~  e—f))] 

=  a-[b-{c-{d-^e+f)}] 
=  a-  [b-{c-d  +  e-f}] 
=  a  —  [b~c  +  d  —  e  +/] 
=  a  —  6  +  c  —  d  +  e  — /. 

45.  Remove  the  marks  of  aggregation  from  the  expres- 
sion 7a —  {3a  —  [4a  —  (5a  —  '^(^)'}},  and  afterwards 
reduce  the  result  to  its  simplest  form.  Also  combine 
and  remove  at  the  same  time.  Result,  5a. 

46.  Kemove  the  marks  of  aggregation  from  a  -f  26  — 
{Qa  —  [36  +  (8.r  —  2  +  6?/  —  x)  +  4a]  —  36}. 

Result,  8b  —  a  +  Ix  —  by  —  2. 
ScH.  3. — Terms  having  common  literal  factors  may  be  regarded  as 
similar  with  respect  to   these,   and  treated    accordingly,   the   other 
factors  in  each  term  being  regarded  as  coefficients. 


*  This,  and  the  examples  wliich  follow  in  this  Bection  may  be   omitted  and  taken 
in  the  review  after  Fractions,  if  thought  best  by  the  teacher. 


SUBTBACTION.  49 

47.  From  ax  -\-  by  —  cz  take  my  —  nx  -\-  2z. 

Result,   {a  +  n)x  -{-  {b  —  m)y  —  (c  +  2)z. 

48.  From  ax'^  -\-  bxy  +  cy^  take  dx^  —  hxy  +  ky~. 

Result,   {a  —  d)x-  +  (6  +  h)xy  -f  (c  —  k)i/-. 

49.  From  v/F+1/  +  3ax  —  12  take  4(^7  +  y)'^ -\- b  —  2ax. 

Result,  5ax  — •  3(^  +  y)"-^  —  12  —  b. 

50.  From  a-x-  ■ — ■  4,axy  -{-4:a-x-y^  take  c^  jt-  —  Sexy  +  Gx-y\ 
Result,   {a-  —  c-)x-^  —  (4a  —  Sc)xy  +  (4a2  —  6)x-^y-. 

Show  that  the  second  term  in  the  last  result  may  be  +  (8c  —  4a). 

51.  From  Vx^^^-  —  2(a  +  ^)  ^+  3  take  - 


4(^._y.)i_l. 


Defi 


Result,   {a  -\-  xy 


Synopsis  for  Review. 

Subtraction. 

Subtrahend. 

Minuend. 

^■a,  \  Scholium. 

Difference,    j  m^^stration.     Diagram. 

^       />  7  r>     t     T>  n  \  Sch.  1.  Both  add.  and  sub'i 

^}^  General  Prob.    Kui.E.     Dem.    |  ^^y^.  2.  Practical  method. 

2  (  Cor.  1. — To  remove. 

g      Brackets.  \  Cor.  2. — To  introduce. 
5^   I  /  Cor.  3.— Several. 

[  Terms  partially  similar.     Sch.  3.  ' 


Test  Questions. — "WTiat  two  answers  can  you  give  to  the  question, 
"  "What  is  the  difference  between  10  and  6  ?"  "Why  do  you  change  the 
signs  of  the  subtrahend  in  subtracting  ?  Why  do  you  add  the  sub- 
trahend, -v^ith  signs  changed,  to  the  miniiend  ?  When  do  you  change 
the  signs  in  removing  a  parenthesis  ?  Why  ?  AVhat  becomes  of  the 
sign  before  the  brackets  ?  In  removing  a  parenthesis  preceded  by  a 
—  sign,  is  the  sign  of  the  first  terci  changed  as  well  as  the  others  ? 


50  FUNDAMENTAL   RULES. 

State  in  the  briefest  manner  the  theorj^  of  subtraction  ?  Iteply.  Sub- 
traction is  finding  the  difference  between  quantities,  that  is,  finding 
what  must  be  added  to  one  qiiantity  to  produce  the  other.  This 
difference  may  always  be  considered  as  consisting  of  two  parts,  one  of 
which  destroys  the  subtrahend,  and  the  other  part  is  the  minuend 
itself.  Hence,  to  perform  subtraction,  we  change  the  signs  of  the 
subtrahend  to  get  that  part  of  the  difference  which  destroys  the  sub- 
trahend, and  add  this  result  to  the  minuend,  which  is  the  other  part 
of  the  difference. 


SECTION  IK 
Multiplication. 

81.  JfllltiplicatiOfl  is  the  process  of  finding  the 
simplest  expression  consistent  with  the  notation  used,  for  a 
quantity  which  shall  be  as  many  times  a  specified  quantity, . 
or  such  a  part  of  that  quantity,  as  is  represented  by  a 
specified  number.  \  The  latter  number  is  called  the  Multi- 
plier. The  quantity  to  be  multiplied  is  called  the  Multipli- 
cand. Taken  together,  the  multiplier  and  multiplicand  are 
called  Factors.     The  result  is  the  Product. 

82.  CoR.  1. — The  multiplier  must  always  he  conceived  as  an 
abstract  number,  since  it  shows  how  many  times  the  multii)li- 
cand  is  to  be  taken.  Thus,  to  proj^ose  to  multiply  $12  by  $5 
is  absurd.  We  can  understand  that  5  times  $12  is  $60  ; 
but  what  is  meant  by  5  dollars  times? 

83.  Cor.  2. — The  jjroduct  is  always  of  the  same  kind  as  the 
multiplicand. 

S-Jt.  ScH. — It  is  frequently  convenient  in  j^ractice  to  speak  of  the 
multiplier  as  positive  or  negative,  although,  literally  understood,  this 
is  a  contradiction  of  Cok.  1,  which  requires  the  multiplier  to  be  con- 
ceived as  mere  number.  In  a  strict  analysis,  the  multiplier  in  such 
cases  is  to  be  considered,  first,  -svithout  reference  to  its  sign,  i.  c,  us 


MULTIPLICATION.  51 

abstract,  and  then  the  sign  is  to  be  interpreted  as  indicating  what  is 
to  be  done  with  the  product,  when  it  is  taken  in  connection  with 
other  quantities. 


SS,  JPvojy*  1  • —  The  product  of  several  factors  is  the  same 
in  whatever  order  they  are  taken. 

Dem. — 1st.    a  X  &,  is  a  taken  h  times,  oxa-\-a-\-a-\-a-\-a 

to  h  terms.  Now,  if  we  take  1  unit  from  each  term  (each  a),  we  shall 
get  h  units  ;  and  this  process  can  be  repeated  a  times,  giving  a  times 
6,  or  5  X  a.  • '  •  aXh  =  hX  a. 

2nd.  When  there  are  more  than  two  factors,  as  ahc.  We  have 
shown  that  ab  ^=zba.  Now  call  this  product  m,  whence  abc  =  mc. 
But  by  part  1st,  mc  =z  cm.  .  • .  ahc  =  hac  =  cctb  s=  cba.  In  like  man- 
ner we  may  show  that  the  product  of  any  number  of  factors  is  the 
same  in  whatever  order  they  are  taken,     q.  e.  d. 

ScH. — If    the    multiplicand  is  concrete,  the  reasoning  is  still  the 

same.      Thus  $a  X  ?>  =  $«  +  $a  +  ^ct  +   ^«> etc.   to  b 

terms.  Now  take  $1  from  each  of  the  terms  of  $a  each,  and  we  have 
$b ;  and  this  process  can  be  repeated  a  times,  giving  $6  X  «•  .  * . 
$a  X  b  =  S&  X  a.  Notice  that  in  each  case  the  multiplier  is 
abstract. 

80,  JPl*op,  2. —  IVhen  two  factors  have  the  same  sign  their 
product  is  positive :  ivhen  they  have  different  signs  their  pro- 
duct is  negative. 

Dem. — 1st.  Let  the  factors  he  -\-  a  and  +  h.  Considering  a  as  the 
multipher,  we  are  to  take  +  6,  a  times,  which  gives  -\-  ab,  a  being  con- 
sidered as  abstract  in  the  operation,  and  the  product,  -\-  ab,  being  of 
the  same  kind  as  the  multipHcand  ;  that  is,  positive.  Now,  when  the 
product,  +  ab,  is  taken  in  connection  with  other  quantities,  the  sign 
-j-  of  the  multipher,  a,  shows  that  it  is  to  be  added ;  that  is,  MTitten 
wi>th  its  sign   unchanged.  .  • .   (-j-  b)  X  (+  <*)  =  +  «&• 

2nd.  Let  the  factors  be  —  a  and  —  b.  Considering  a  as  the  multi- 
pher, we  are  to  take  —  b,  a  times,  which  gives  —  ab,  a  being  consid- 
ered as  abstract  in  the  operation,  and  the  product,  —  ab,  being  of  the 
same  kind  as  the  multiphcand  ;  that  is,  negative.  Now,  when  this 
product,  —  ab,  is  taken  in  connection  mth  other  quantities,  the  sign 
—  of  the  multipher  shows  that  it  is  to  be  subtracted  ;  that  is,  writ- 
ten with  its  sign  changed.     .  • .  { —  6)  X  (  —  a)  =  -|-  a6. 


52  FUNDAMENTAL   KULES. 

3d.  Let  the  factors  be  —  a  and  -f-  ^-  Considering  a  as  the  multi- 
pUer,  we  are  to  take  -{-b,  a  times,  which  gives  -{-ab,  a  being  considered 
as  abstract  in  the  operation,  and  the  product,  -|-  ab,  being  of  the  same 
kind  as  the  multiplicand  ;  that  is,  positive.  Now,  when  this  product, 
+  ab,  is  taken  in  connection  with  other  quantities,  the  sign  —  of  the 
multiplier  shows  that  it  is  to  be  subtracted  ;  that  is,  written  with  its 
sign  changed.    .  • .   (+  ^)  X  ( —  a)  =  —  ab. 

4th.  Let  the  factors  be  -f-  «  and  —  b.  Considering  a  as  the  multi- 
pher,  we  are  to  take  —  b,  a  times,  which  gives  —  ab,  a  being  consid- 
ered as  abstract  in  the  operation,  and  the  product,  —  ab,  being  of  the 
same  kind  as  the  multiplicand  ;  that  is,  negative.  Now,  when  this 
product,  —  ab,  is  taken  in  connection  with  other  quantities,  the  sign 
-|-  of  the  multiplier  shows  that  it  is  to  be  added  ;  that  is,  written  with 
its  own  sign.     .  •.   ( —  Z/)  x  (-f-  ^)  =  —  o^-     Q-  e.  d. 

87 •  CoE.  1. —  The  product  of  any  number  of  positive  fac- 
tors is  positive.  Thus  (+  a)  X  (+  6)  X  (+  c)  x  (-\-d)  = 
abed,  since  (+  a)  X  (+  b)  =  -\-ab,  which,  in  turn  multiplied 
^y  +  c,  gives  +  abc,  &c. 

88,  CoK.  2. — The  product  of  an  even  number  of  negative 
factors  is  positive  ;  since  ive  can  multiply  them  two  and  two, 
thus  obtaining  positive  products,  which  positive  products  multi- 
plied together  make  the  complete  product  positive.  Thus 
(  -  a)  X  (  -  6)  X  (-  c)  X  C  -  ^)  X  (  -  e)  X  (  -/) 
z=  (  -\-  ab)  X  i  -\-  cd)  X  {  -\-  ef )  =  -\-  abcdef  or  abcdef 

89,  Cor.  3. — The  product  of  an  odd  number  of  negative 
factors  is  negative ;  since,  by  the  last  corollary,  the  product  of 
all  but  one  {an  even  number)  of  such  factors  is  positive,  and 
then  this  multiplied  by  the  remaining  negative  factor  gives 
-f  X  —  ,  and  hence  is  negative. 


00,  I^vop,  3, —  TJie  product  of  two  or  more  factors  con- 
sisting of  the  same  quantity  affected  with  exponents,  is  the  conn- 
mi  m  quantity  with  an  mponent  equal  to  the  sum  of  the  expO' 


MULTIPLICATION.  53 

nents  of  the  factors.  That  is  a"  x  a"  =  «""+";  or  a"" ■a'' a* 
__  ^m-vn+,^  ^^^  whether  the  exponents  are  integral  or  frac- 
tional, positive  or  negative. 

Dem. — 1st.  When  the  exponents  are  positive  integers.  Let  it  be 
required  to  multiply  a'-^  by  a^.  a'-^  =  aaa,  and  a*  =  aa.  .  ••  a^  X  a* 
=  aaa-aa  =  a^.  That  is,  there  are  tliree  factors  each  a,  in  a^,  and  tico 
like  factors  in  a^ ;  and,  as  the  product  consists  of  all  the  factors  in 
both  multiplier  and  multiplicand, it  will  contain^re  factors  each  a,  and 
hence  is  a^.     In  general :   To  multiply  a"*  by  a"  mid  a',     a"   =  aaaa 

to  rn  factors,  a"  =  aaaaa to  n  factors,  and  a'   =  aaaaa 

to  s  factors.     Hence  the  product,   being  composed  of  all  the 

factors  in  the  quantities  to  be  multiplied  together,  contains  77i  -f-  ^^  ~f" 
s  factors  of  a,  which  is  expressed  a*"  +  "  + «,  Since  it  is  evident  that 
this  reasoning  can  be  extended  to  any  nxmiber  of  factors,  as  a""  X 
a"  X  a*  X  cf  ,  etc.,  etc.,  the  proposition  in  this  case  is  proved. 

2nd.    When  the  exponents  are  positive  fractions.     Let  it  be  required 

to  multiply  64^"  by  64^.     Now 64^  =  4-  4,  i.  e.,  2  of  the  3  equal  fac- 

tors  which  make  64.    In  like  manner  64'' is2    2  •  2  •  2  -  2.    Andsince4-4 

is2  -  2  -  2  -  2,  64^  X  64^  =  2  •  2  .  2  •  2  X  2  -  2  ■  2  •  2  -  2  =  2^  or  9  of  the  6 

.p. 
equal  factors  into  which  64  is  resolvable,  and  may  be  expressed  64  ^   .  • . 

64^  X  64''  =  29  =  64  *  =64^    **.      In  general,  let  a "  be  multiphed  by 

a*,  a"  means  m  of  the  n  equal  factors  composing  a.  Now,  if  each  of 
these  71  factors  be  resolved  into  h  factors,  a  will  be  resolved  into  6n 

m 

factors,   and  to  make  the    quantity  a»  we   shall    have  to    take    hm 

m  6m  e 

instead  of  m  factors.     Hence  a"  =  a^".     In  like  manner  a*  may  be 

en  m  e  bm  en 

shown  equal  to  a'"' ;  and  a"^  x  a^  =  «''"  x  a^'\  This  now  signifies 
that  a  is  to  be  resolved  into  b)i  factors,  and  bm  +  en  of  them  taken  to 

m  e  bm  en  bm+en  »i      <• 

form  the  product,  .'.a"-  x  a*^  =  «*"  x  «*"  =  «  *»  ,  or  a"  *^  which 
proves  the  proposition  for  positive  fractional  exponents,  since  the 

m  < 

same  reasoning  can  be  extended  to  any  number  of  factors,  as  a**  x  a* 

X  a\  etc. 
3d.  When  tJie  exponents  are  negative.    Let  it  be  requii-ed  to  multiply 


54  FUNDAMENTAL  EULES. 


2-3    by  2  -  2 .      2-3    is  -— ,  and  2  -  2  is-j     .  • .     2-3x2-2=:; 

111  11 

-rj  X  t;!  —  IT?,  or  2  -s      So  also  a-*"  Xa-"  =  —  X  — .    Now, 
2  2  2  a'"       a" 

as  fractions  are  multiplied  by  multiplying  numerators  together  and 

denominators  together,  we  have  —  X  —  = r-  by  part  1st  of  the 

&  a'"        a"        a"'  +  "     "^  -^ 

demonstration.     But  this  is  the  same  as  a  ~  <''*  +  "^  or  a  -  ™  -  "    .  • . 

--  -^       1  1  1 

a -'"Xa- »  =  «-"*-".    Finally  a      »Xa    ''=-;;^   ><  ^7=  ~; — ~ 

a"^        a*        a»    V 

a      n   "«>  .     Q.  E.  D. 

EXAMPLES. 

1.  Prove  as  above  that  81^   x  81*  =  81^  and  that  81^^ 

9. 

=  81*. 

2.  Prove  that  m"  x  m*  :=m''+*. 

3.  Prove  that  16"*  x  16""^  =  16~^. 

4  Prove  that  25  ~  ^  x  25^  is  1. 
5.  Prove  that  a~'  x  a^  is  a. 

ScH. — The  student  must  be  careful  to  notice  the  difference  be- 
tween the  signification  of  a  fraction  used  as  an  exponerit,  and  its  com- 
mon signification.  Thus  |  used  as  an  exponent  signifies  that  a  num- 
ber is  resolved  into  3  equal  factors,  and  the  product  of  2  of  them 
taken  ;  whereas  f  used  as  a  common  fraction  signifies  that  a  quantity  is 
to  be  separated  into  3  equal  ^«rte,  and  the  sum  of  two  of  them  taken. 


01,  I*TOh, — To  multiply  monomials. 

R  ULE. — Multiply  the  numerical  coefficients  as  in  the 

DECIMAL  NOTATION,  AND  TO  THIS  PKODUCT  AFFIX  THE  LETTERS 
OF  ALL  THE  FACTORS,  AFFECTING  EACH  WITH  AN  EXPONENT 
EQUAL  TO  THE  SUM  OF  ALL  THE  EXPONENTS  OF  THAT  LETTEB 


MULTIPLICATION.  55 

IN  ALL  THE  FACTORS.  ThE  SIGN  OF  THE  PRODUCT  WILL  BE 
+  EXCEPT  WHEN  THERE  IS  AN  ODD  NUMBER  OF  NEGATIVE  FAC- 
TORS ;     IN    WHICH    CASE   IT   WILL   BE   . 

Dem.— This  rule  is  bjit  an  application  of  the  preceding  principles. 
Since  the  product  is  composed  of  all  the  factors  of  the  given  factors, 
and  the  order  of  arrangement  of  the  factors  in  the  product  does  not 
affect  its  value,  we  can  write  the  product  putting  the  continued  pro- 
duct of  the  numerical  factors  first,  and  then  grouping  the  hteral  fac- 
tors, so  that  hke  letters  shall  come  together.  Finally,  performing  the 
operations  indicated  by  multiplying  the  numerical  factors  as  in  the 
decimal  notation,  and  the  like  literal  factors  by  adding  the  exponents, 
the  product  is  completed. 

EXAMPLES. 

1.  Multiply  together  Sa'^-bx,  2cb^y,  and  5ac^x. 

Model  Solution. — Since  the  product  must  contain  all  the  factors  of 
the  given  factors,  and  the  order  of  arrangement  is  immaterial  (85), 
3a2  bx  X  2c6^y  X  ^ac^x  =  S  ■  2  ■  5  X  a^a  X  hh^  X  cc^  X  xx  X  y, 
which,  by  performing  the  operations  indicated,  becomes  dOa'^b^c^x^y. 

2.  Multiply  together  dabxy,  2a*bx\  lOcx^,  ^y\  and  a. 

Prod.,  2^0d'¥cx^y\ 

3.  Multiply  together  5ax%  —  2by,  —  Sa^c,  a^,  and  —  2y^. 

Prod.j  —  QOa^hcx''>y\ 

4.  Multiply  together  m^",  ^rrV^x^,  —  3a^^  —  5m',  and  ^a^x\ 

Prod.,  120a^m'  +  V +  •"  +  ". 

5.  What  is  the  product  of  —  2c"'ci,  —  lOac",  —d^x,  —  4a'"ar", 
aud  —  c\  Ans. ,  —  80a"*  -^  'c"*  + "  +  V^^"  +  \ 

6.  Multiply  Zx^  by  2x^.  Prod.,  Qx^. 

7.  Multiply  60a^  by  8a^.  Prod.,  480a«. 

8.  Multiply  ^ah^  by  —  lah^.  Prod.,  —  21ah^, 


56  lUNDAMENTAL   RULES. 


9.  Multiply  5ah^  by  la^b-^c. 

Prod.,  ^5ah. 

10.  Multiply  10a'  by  3a -2. 

Prod.,  30. 

11.  Multiply3i/"by  6t/2. 

1  +  2h 

•  Prod.,  182/~~. 

1            1 

12.  Multiply  13^"  by  5x"\ 

W  +  n 

Prod.,  65^  '""  . 

1                5 
13.  Multiply  —  50a'  by  —  ^a\ 

6 

Prod.,  200a". 

14.  Multiply  309a'^6'»  by  9a™6". 

3 
Pro6?.,  2781a'"6'"  +  ». 

15.  Multiply  —  5^  -  V  - '  by  — 

2^x^y\             Prod.,  Ibxy. 

16.  Multiply  ^^  by  ^  ~  ^.. 

Prod.,  x^. 

17.  Multiply  a:™  by  a;  -  "*. 

Prod.,  1. 

18.  Multiply  a^b"  by  a^b'. 

Prod.,  aK 

19.  Multiply  v'o^'by  a"^x. 

Prod.,  a^'x^. 

20.  Multiply  V^by  v^^ 

Prod.,  a^. 

02,  I*VOh,  To  multiply  two  factors  togetKer  when  one  or 
both  are  polynomials. 

RULE. — Multiply   each  term    of    the   multiplicand   by 

EACH  TERM  OF  THE  MULTIPLIER,  AND  ADD  THE  PRODUCTS. 

Dem. — Thus,  if  any  quantity  is  to  be  multiplied  by  a  -{-h  —  c,  if  we 
take  it  a  times  (i.  e.  multiply  by  o),  then  h  times,  and  add  the  results, 
we  have  taken  it  a  -j-  ^  times.  But  this  is  taking  it  c  too  many  times, 
as  the  multiplier  required  it  to  be  taken  a  -\-h  minus  c  times.  Hence 
we  must  muliply  by  c,  and  subtract  this  product  from  the  sum  of  the 
other  two.  Now  to  subtract  this  product  is  simply  to  add  it  with  its 
signs  changed  {77)'  But,  regarding  the  —  sign  of  c  as  we  multiply, 
will  change  the  signs  of  the  product,  and  we  can  add  the  partial  pro- 
ducts as  they  stand,  even  without  first  adding  the  products  by  a  and 

h.      Q.  E   D. 


MULTIPLICATION.  57 


EXAMPLES. 

1.  Multiply  2a  —  36  +  4c  by  3a  +  26  —  5c. 

Model  Sol-dtion. — Writing  the  multiplier  under  the  multiplicand, 
as  a  matter  of  convenience,  I  have 
2a  —  36   +  4c 
3a   +  2b   —  5c 
6a2  —  dab  +  12ac 

+  4a&  —  66*  +    86c 

-  lOac     +  156c  —  20cg 

6a^  —  dab+   2ac  —  06^=  -f  236c  —  2Uc=* 

Novr  taking  the  multiplicand  3a  times,  I  have  6a*  —  9a6  -j-  12ac. 
Taking  it  26  times,  I  have  4a6  —  66*  -|-  86c.  I  have  thus  taken  it  too 
many  times,  by  5c  times.  Hence  I  am  to  take  it  5c  times,  and  then 
subtract  this  partial  product  from  the  others.  Therefore  I  multiply 
by  5c,  and  change  the  signs  as  I  proceed,  and  finally  add  the  three 
partial  products.     I  thus  obtain  3a  -j-  26  —  5c  times  the  multiplicand. 

2.  Multiply  X  -\-  y  hy  X  -{-  y.  Prod.,  x'^  -f  2xy  +  y^. 

3.  Multiply  5x  +  4?/  by  3x  —  2y. 

Prod.,  15^2  _|_  207?/  —  8?/2. 

4.  Multiply  x'^  -\-  xy  —  y^  hj  x  —  y. 

Prod.,  x^  —  2xy^  +  ?/3. 

5.  Multiply  2ac2  —  dby  by  2c^  —  3y'K 

Prod.,  4ac5  —  Qbc^y  —  GacHj'^  4-  dbifo 

6.  Multiply  a*  _{-  62  _|_  ^s  —  ab  —  ac — be  by  a  4-  b  +  c. 

7.  Multiply  a^  -f-  da^x  +  Sax""  +  x^  by  a^  —  Sa'^x  +  Sax^ 

x\  Prod.,  a'—  3a*x^  +  Sa'^x*—  x\ 

8.  Multiply  a""  —  V"  by  a""  +  l"". 

I  IT  III 

9.  Multiply   together    {a  +  1)),    {ci'^  +  ah  +  V"),  {a  —  h\ 

IV 

and  {a"^  —  nh  +  If^). 


58  FUNDAMENTAL  RULES. 

Suggestion. — Perform  this  first  by  multiplying  together  I  and  II, 
and  men  III  and  IV,  and  taking  the  product  of  these  products.  2d, 
By  multiplying  the  product  of  I  and  III  into  the  product  of  II  and  IV. 
3d,  By  multiplying  the  product  of  I  and  IV  by  the  product  of  11  and 
III.  Besult  in  each  case,  a*"  —  h^. 

10.  Multiply  a3  —  ^3  by  m^  —  w^. 

Frod.,  a/m- —  tr-x^  —  a^n^  +  n-x^. 

11.  Multiply  2fl2  -{-  2a  +  5  by  a"-  —  a. 

Prod.,  2a*  +  Sa^  —  5a. 

12.  Multiply  2^3  4.  4^2  _^  g^  ^  ig  by  3^  —  6. 

Prod.,  6^^  —  96. 

13.  Multiply  a  -\-h  —  c  by  m  —  n. 

Prod.,  am  —  an  -\-  hm  —  hyi  —  cm  -f-  en. 

14.  Multiply  X*  -\-  x^y^  +  y*  by  ^2  —  y\ 

Prod.y  x^  —  2/6. 

15.  Multiply  a;2  —  4^  +  16  by  J7  +  5. 

Prod.,  x^  -\-  x'^  —  ^x  ^  80. 

16.  Multiply  a*  —  a^y  +  a^y^  —  ay^  -f  y*  by  a  +  y. 

Prod.,  a^  +  y\ 

17.  Multiply  ^2  _  mx  —  100  by  ^  +  2. 

Prod.,  x^  —  48-2;2  —  200^  —  200. 

'    18.  Multiply  2x-^  +  3^  —  1  by  2j;2  _  3j:  +  1. 

Prod.,  4^^  —  9a;2  +  6ir  —  1. 

19.  Multiply  j:^'  —  h^  by  jt'-^  +  6. 

Prod.,  x^  —  h^x^-]-hx^  —  6^. 


7  _  2       1  .i 


20.  Multiply  d^  —  h     ^  hy  a^  —  h 

Prod.,  a^  —  a'^h  ~  ^— a"^  b  +  6^. 

21.  Multiply  2a~^  —  36^  by  2a  "^  +  36^. 

Prod.,  4a  ~  '  —  96^. 


MULTIPLICATION.  59 

93333  9  3  3 

22.  Multiply  X*  -f  x^y  ~  *  -f  x^y  '^-^  y'^  hj  x^ — y  " ^. 

Frod.,  x'^  —  y~^' 

23.  Multiply  a"'  -f  6"'  by  a"  +  6". 

1  i  ^  1  Q  2 

24.  Multiply  ^"^  +  2/^  by  ^'^ — y^-  Prod.,  xr^ —  ij^. 

2.  11  3  1  1 

25.  Multiply  m^  +  ??i^n^  -f  ri^  by  ?7i^  —  n^. 

Prod.,  m  —  n. 

26.  Multiply  Sa™  -  ^  _  2&" "  ^  by  2a  —  S?;^. 

Prod.,  Ga"*  —  4a6"- '  —  9a"' "  '6^  +  66". 

27.  Multiply  2a"'  -^6  ~  =^  +  a  "  '^6^  by  da^V  —a'^b-  '^. 

P?'oc?.,  Ga^'"  -p  -{-  3a"' "  '^6'^  —  2a"'  +  2^6"  '^'  —  a^'b  -". 

^5.  Def.  — Wlien  an  indicated  operation  is  performed  the  expres- 
sion is  said  to  be  expa^ided. 

28.  Expand  (a  +  6)(a  —  6)  ;  also  {x  -f  ?/)(^  +  y)  ;  also 
(^  —  a) {x^ -\-ax -{- a-) ;  also  (7?i  -\-  n){7n  -\-  n)  —  (m  —  ?i) 
(w  —  n).  Last  Result,  4r)in. 


THREE  IMPORTANT  THEOREMS. 

04,  Theorem. — The  square  of  the  sum  of  two  quantities 
is  equal  to  the  square  of  the  first,  plus  twice  the  product  of  the 
two,  plus  the  square  of  the  second. 

Dem. — Let  a;  be  any  one  quantity  and  y  any  other.  The  sum  is 
X  -\-  y  ;  and  the  square  is,  the  square  of  the  first,  x^,  plus  twice  the 
product  of  the  two,  2xy,  plus  the  square  of  the  second,  y^.  That  is 
(a;  +  2/)'  =x^  -h  2a-y  +  y^.  For  {x  +  yY  =  {x  -\-  yXx  +  y) 
which  expanded  becomes  x*  -\-  2xy  -\-  y^.     Q.  e.  d. 

EXAMPLES. 

1.  What  is  the  square  of  2a2  +  3x^  ? 


60  FUNDAMENTAL  EULES. 

Model  Solution. — This  is  the  sum  of  the  two  quantities  2rt*  and 
3a;*,  and  hence  by  the  theorem  the  square  equals  the  square  of  the 
first,  4a'*,  plus  twice  the  product  of  the  two,  12a^x^,  plus  the  square 
of  the  second,  djo\     .-.     {2a^  +  3x2)^  =  4a*  +  12a''x^  +  9x\ 

ScH. — The  pupil  should  give  mentally  the  squares  of  the  follow- 
ing  expressions : 

2.  Square  4a^  +  2x.  liesuU,  16a  +  IQ^c^x  +  Ix'K 

_  2.  _  ^  1 

3.  Square  x    ^  -\-  xy.  Besult,      ^    •*  +  2x'^y  +  x'-y\ 

4  Square  5a  -  '  +  3a'^»^ 

Hesult,   25a-^  -{-  SOat^  +  da^b*. 
5.  Square  |a6  -  ^  _j_  f  j?  -'  6  -  •. 

Hesult,  ^a^b -'^  -\-  ^ab~"^x-^  -{-  t-^-^b -\ 


0^»  Theo. — Ulte  square  of  the  difference  of  two  quanti- 
ties is  equal  to  the  square  of  the  first,  ininus  twice  the  product 
of  the  first  by  the  second,  plus  the  square  of  the  second. 

Dem. — Let  a;  and  y  be  any  two  quantities.  The  difference  is 
jc  —  y.  Now  (X  —  yY  =  (ic  —  y){x  —  y)  which  expanded  gives 
jc2  —  2xy  ■\-  y^.     Q.   E.   D. 

EXAMPLES. 

1.  Square  2x  —  By.  Resvdi,  Ax^  —  12xy  +  dy^. 
Model  Solution. — Similar  to  the  last 

2.  Square  x~'^y  —  2y'.     Result,  x~^y^  —  A:X~'^y^  -J-  Ay*. 

m  _1 

3.  Square  2a^  —  36  -. 

2m  ml  J 

Result,  4a"'  —12a  "^6"""+  96~"". 

4.  Square  m~^ — w~'. 

Result,  m-""  —  2m --^n"'  +  ?2-^ 

5.  Square  3a   "*  —  2b    " . 

2  1  m  2nt 

Resu.'t,   9a    ^ —  12u    mlf~  n  -\-  4:b~  "«  . 


MULTIPLICATION.  61 

00*  Theo. — Tlie  product  of  the  sum  and  difference  of 
two  quantities  is  equal  to  the  difference  of  their  squares. 

Dem. — Let  X  and  y  be  any  two  quantities.  Their  sum  is,  x  -\-  y,  and 
their  difference  is  x  —  y.  Now  (x  +  y)  multipUed  by  (x  —  y)  gives, 
by  actual  multiplication,  x*  —  y^,  or  the  difference  of  the  squares  of 
the  two  quantities,     q.  e.  d. 

EXAMPLES. 

1.  Find  the  product  of  2a^  +  36  and  2a2  —  36. 

Model  Solution. — Here  I  have  the  sum  of  the  two  quantities  2a* 
and  3h,  and  their  difference.  Therefore  {2a^  +  3b)  X  (2a*  —  36)  is, 
by  the  theorem,  the  square  of  2a*,  or  ia'',  minus  the  square  of  the 
second,  or  9b-.  .-.     (2a*  X  36)  (2a*  —  36)  =:  4a*  —  96*. 

2.  Find  the  product  of  a  +  26  and  a  —  26. 

Frod.,  a-^  —  462. 

3.  Find  the  product  of  2a  +  36  and  2a  —  36. 

Prod.,  4a^  — 962. 

4.  Find  the  product  of  la  +  26  and  la  —  26. 

Frod.,  49a2  — 462. 

5.  Find  the  product  of  5a^  +  662  and  5a^  —  66^. 

Frod:,   25a<^  —  366^ 

11  11 

6.  Find  the  product  of  m^  +  n^  and  m^  —  n'^. 

Frod.,  m  —  n. 

7.  Find  the  product  of  2^x^  +  S^y^  and  2^x^  —  S^y^, 

Frod.,  2x  —  3?/. 


Frod.,  da'¥  —  ^ab\ 


8.  Find  the  product  of  30^63  +  2a26^  and  3a-'6a  —  2a^bK 

X 


82 


FUNDAMENTAL   KULES. 


Synopsis  for  Eeview. 


Defs.  i 


[Multiplication.  "] 
Multiplier. 
Multiplicand 
Product. 
Factors. 


-    Corollaries. 


Fundamental 
Propositions. 


1st.  Order  of 
Factors.    Dem 


2d.  Law  of 

Signs.  Dem. 


3d. 


Laws  of  expo 
nents.    Dein 


'  1.  Multiplier,  ab- 
stract. 
2.  Product     like 
J  multiplicand. 

"sGH.How  the  sign 
of     a  multi- 
plier is  to  be 
(.  understood. 

Two  factors,  ) 
More     than  V  Scho. 


two. 


Give  + 


Give 


Cor's. 


1.  +  Fact's. 

2.  Even  No. 

—  Fact's. 

3.  Odd   No. 

—  Fact's. 


To  perform  Multipli- 
cation. 


Three  important 
Theorems. 


Prob.  2. 


Positive  integer. 
Positive  fraction. 

Negative.    I^^^^f^' 
^  ( Fraction. 

What?    EuLE.     De7n. 

What  ?    KuLE.     I)e7?i. 


Square  of  sum.     Dem. 
Square  of  difference.     Dem. 
Product  of  sum  and  difference. 


Dem. 


Test  Questions. — State  and  demonstrate  the  law  of  the  signs  in 
multiplication.  How  are  expressions  consisting  of  the  same  quantity- 
affected  by  exponents,  multiplied  ?  How  many  cases  arise  ?  Demon- 
strate each.  What  is  the  difference  between  the  square  of  the  sum, 
the  square  of  the  difference,  and  the  product  of  the  sum  and  differ- 
ence of  two  quantities  ?    Demonstrate.    What  is  the  product  of  a'-^ 


and  a"*  ?    Prove  it. 

2. 

and  a  ^  ?    Prove  it. 


What  of  a 
What  of  a' 


^  and  a  -  2  ? 
»  and  a  -  "»  ? 


Prove  it.     What  of  a' 
Prove  it. 


DIVISION.  63 

SECTION'   V. 

Division. 

97*  Division  is  the  process  of  nnding  how  many 
times  one  quantity  is  contained  in  another.  The  JDivi' 
dend  is  the  quantity  to  be  divided.  The  Divisor  is  th3 
quantity  by  which  we  divide.  The  Quotient  is  the  result, 
which  shows  how  many  times  the  divisor  is  contained  in 
the  dividend.  The  JRemaindev  is  what  is  left  of  the 
di^ddend  after  the  integral  part  of  the  quotient  is  pro- 
duced. 

98.  ScH.  1. — Division  is  the  converse  of  multiplication.  Since 
a  product  consists  of  (contains)  as  many  times  the  multiplicand  as 
there  are  units  in  the  multipUer,  the  multiplier  shows  how  many 
times  the  multipHcand,  is  contained  in  the  product.  The  product, 
therefore,  corresponds  to  the  dividend,  the  multiplicand  to  the  divi- 
sor, and  the  multipher  to  the  quotient.  But,  as  in  multiplication, 
multiplicand  and  multipher  may  change  places  without  affecting  the 
product,  either  of  them  may  be  considered  as  divisor  and  the  other  as 
quotient,  the  product  being  the  dividend. 

99,  ScH.  2. — In  accordance  with  the  last  schohum,  the  problem 
of  division  may  be  stated  :  Given  the  product  of  two  factors  and  one  of 
the  factors,  to  find  the  other ;  and  the  sufficient  reason  for  any  quotient  is, 
that  multiplied  by  the  divisor  it  gives  the  dividend. 

100,  Cor.  1. — Dividend  and  divisor  may  both  be  multi- 
plied or  both  be  divided  by  the  same  number  without  affecting 
the  quotient. 

This  truth  is  axiomatic,  but  may  be  illustrated  thus  :  If  a  given 
number  of  apples  are  divided  among  any  number  of  boys,  each  boy 
will  receive  just  the  same  number  as  if  twice  or  thrice  .as  many  were 
divided  among  twice  or  thrice  as  many  boys,  or  as  if  ^  or  ^  as  many 
were  divided  among  ^  or  ^  as  many  boys. 

101,  Cor.  2. — If  the  dividend  be  multiplied  or  divided  by 

any  number,  whih  the  divisor  remains  the  same,  the  quotient  is 
multiplied  or  divided  by  the  same. 


64  FUNDAMENTAL   BULES. 

This,  too,  is  an  axiom,  but  may  be  illustrated  as  the  last. 

102,  Cor.  3.  If  the  diviHor  he  multiplied  by  any  number 
while  the  dividend  remains  the  same,  the  quotient  is  divided  by 
that  number  ;  but  if  the  divisor  be  divided,  the  quotient  is  mul- 
tiplied. 

This,  like  the  two  preceding,  is  an  axiom,  but  may  be  illustrated  in 
a  similar  manner. 

103 •  Cor.  4.— The  sum  of  the  quotients  of  tvx)  or  more 
quantities  divided  by  a  common  divisor,  is  the  same  as  the 
quotient  of  the  sum  of  the  quantities  divided  by  the  same 
divisor. 

This  is  an  axiom.  To  illustrate  it,  consider  that  as  2  goes  into  8,4 
times,  and  into  6,  3  times,  it  will  go  into  8  and  6,  or  8  -f-  6,  4  times  -f- 
3  times,  or  7  times,  etc. 

104,  Cor.  5. — The  difference  of  the  quotients  of  two  quan- 
tities divided  by  a  common  divisor,  is  the  same  as  the  quotient 
of  the  difference  divided  by  the  same  divisor. 

This  is  also  an  axiom  which  can  be  illustrated  as  the  last. 

105.  Def. — Cancellation  is  the  striking  out  of  a  factor  common 
to  both  dividend  and  divisor,  and  does  not  affect  the  quotient,  as 
appears  from  {100). 

106,  Lemma  1. —  When  the  dividend  is  positive  the  quotient 
has  the  same  sign  as  the  divisor;  but  when  the  dividend  is  neg- 
ative, the  quotient  has  an  opposite  sign  to  the  divisor. 

Dem. — This  proposition  is  a  direct  consequence  of  the  law  of  the 
signs  in  multiplication,  since  the  dividend  corresponds  to  the  pro- 
duct, and  a  positive  product  arises  from  like  signs  in  the  factors,  and 
a  negative  product,  from  unlike  signs. 

107,  Lemma  2. —  When  the  dividend  and  divisor  consist 
of  the  same  quantity  affected  by  exponents,  the  quotient  is  the 
common  quantity  ivith  an  exponent  equal  to  the  exponent  in  the 


DITISION.  65 

dividend,  minus  that  in  the  divisor.  That  is,  a"*  -^  a"  =  a""  ~ ", 
whether  m  and  n  be  integral  or  fractional,  positive  or 
negative. 

Dem. — This  is  an  immediate  consequence  of  the  law  of  exponents  in 
multiplication,  since,  in  the  corresponding  case  the  exponent  of  the 
product  was  found  to  be  the  su7n  of  the  exponents  of  the  factors. 
'Now,  as  the  dividend  is  the  product  of  the  divisor  and  quotient,  it  fol- 
lows that  the  exponent  of  the  quotient  is  the  exponent  of  the  divi- 
dend minus  that  of  the  divisor,    q.  e.  d. 

EXAMPLES.  ^ 

1.  Divide  a'  by  a^. 

Model  Solution,  a''  -^  a^  =  a*  ;  since  a'^  X  a^  =a'',  and  di^^sion 
is  finding  a  factor  which  multiplied  into  the  divisor  produces  the 
dividend  {99). 


2. 

Divide  w*  by  ni^. 

Quot,  m"r^^ 

3. 

Divide  n"  by  n~\ 

Quot.,  ?i"       or  ?i   "   . 

4. 

1 
Divide  (a6)'"'  by  {aby  . 

Quot., 

(aby^-n  oT{ab)    »    . 

5. 

Divide  a^  by  a\ 

Quot.,  a~^  or-  . 
a- 

6. 

Divide  a  " '  by  a^ 

QuoL,  a~^ov-  . 
a-' 

7. 

Divide  x     ^  hj  x~\ 

Quot.,  x^. 

8. 

Divide  x~^hj  x    ^. 

Quot.,  X  ^  or  — . 
x'^ 

108.  CoR.  1. — Any  quantity  with  an  exponent  0  is  1, 
since  it  may  be  considered  as  arising  from  dividing  a  quantity 
by  itself.  Thus  ^o  may  be  considered  as  x'"  -^  ^"^  which  is  1, 
because  dividend  and  divisor  are  equal.  But  by  the  law  of 
exponents  x"^  -^  x"^  =  x'^  . ' .  x'^  =  1.  In  like  manner  the 
significance  of  any  quantity  with  an  exponent  0  may  be  ex- 


6Q  FUNDAMENTAL   liULES. 

52  52 

plained.     5"  =  1,  for  -—  =  1,  and  also  — -  =  5"  .* .  5°  =  1. 
5  0" 

So,  |1  =  1  or  8";  or  |^  =  1  or  8«,  etc. 

100,    CoR.   2. — Negative   exponents   arise  from   division 

ivhen  there  are  more  factors  of  any  nurnber  in  the  divisor  than 

in  the  dividend.      Thus   a^  _i_  a^  ==   a^-^  {107)  =   a~'^. 

A      •  a^         1  ,       1  ,  .  ^ 

Ai^am  a^ -^  a^  ^=  —  =  —   .*.  a~= — »    wnich    accords 

°  a^        a^  a-i 

with  the  definition  of  negative  exponents  {4:3). 

110,  CoR.  3. — A  factor  may  be  transferred  from  dividend 
to  divisor  {or  from  numerator  to  denominaior  of  a  fraction^ 
which  is  the  same  thing),  arid  vice  versa,  by  changing  the  sign 

1 

of  its  exponent.     Thus  -7—  =  —r-  >  for  - —  =  -—  =  — 
•^  ^  63  a-b^  63  63  a* 

-27-3        •  «~' 

a  ^6    ,   since   - —  =: 
62 


63   = 

a'b-^' 

Thus 

also  -; — 

63 

^xi- 

;  and 

1 

63 

■  =  b- 

-3 

EXAMPLES. 

.  Show  that 

a' 

X 

-%2 

62.^2 

a^-y^ 

Model  Solution.— Since  a-2=:-— and  x-^= — ,   I  have 


cc2  h'X^ 


a-      _  a-^  _  &2    .    y3  ^  52         x'i  _  6-x2 


x2 


—  1    _  2 

2a    '-^x    ^?/ 2???/ 


2.  In  like  manner  show  that    ^ 


*  It  is  assiiiued  that  the  pupil  has  learned  in  Arithmetic  how  to  multiply  and 
divide  fractions,  or  reduce  complex  fractions  to  simple  ones. 


DIVISION.  67 

3.  Free  ,,   _., from  negative  exponents  and  explain 

ha   'x-ij    'z  ^ 

the  process. 


Ill,  I^vdb,  1,  To  divide  one  monomial  by  another. 

RULE. — Divide  the  numerical  coEmciENT  of  the  divi- 
dend BY  THAT  OF  THE  DIVISOR  AND  TO  THE  QUOTIENT  ANNEX 
THE  LITERAL  FACTORS,  AFFECTING  EACH  'VN^TH  AN  EXPONENT 
EQUAL  TO  ITS  EXPONENT  IN  THE  DIVIDEND  MINUS  THAT  IN  THE 
DIVISOR,  AND    SUPPRESSING  ALL  FACTORS  WHOSE   EXPONENTS  ARE  0. 

The   sign   of  the   quotient  will  be  -J-  when  dividend  and 

DIVISOR     have     like    SIGNS,    AND    WHEN     THEY    HAVE    UNLIKE 

SIGNS. 

Dem. — The  dividend  being  the  product  of  divisor  and  quotient,  con- 
tains all  the  factors  of  both;  hence  the  quotient  consists  of  all  the  fac- 
tors which  are  found  in  the  dividend  and  not  in  the  divisor.  Or,  the 
correctness  of  the  rule  appears  from  the  fact  that  it  is  the  converse 
of  the  corresponding  operation  in  multiphcation,  so  that  quotient  and 
divisor  multiphed  together  produce  the  dividend.  The  law  for  the 
sign  of  the  quotient  is  demonstrated  in  (106).     Q.  e.  d. 

EXAMPLES. 

1.  Divide   12a^x'*b  by  3a^x^. 

MODEL  SOLUTION. 

oPEBATioN.       3a-X')12a^x^b 
4ax'-6 

Explanation. — In  the  divisor  there  is  a  factor  3,  hence  there  must 
be  a  factor  4  in  the  quotient,  to  produce  12  in  the  dividend.  So,  also, 
since  there  are  3  factors  of  a  in  the  dividend,  and  2  in  the  divisor, 
there  must  be  1  in  the  quotient  in  order  that  the  product  of  divisor 
and  quotient  may  be  the  dividend.  In  like  manner  4  factors  of  x  in 
the  dividend,  and  2  in  the  divisor,  require  2  in  the  quotient.  There 
being  1  factor  h  in  the  dividend,  and  none  in  the  divisor,  one  factor  of 
5  must  appear  in  the  quotient.  Hence  12a^x^h  -^  da-x-  =  4a.i--&, 
which  quotient  containing  all  the  factors  of  the  di^ddend  not  found 
in  the  divisor,  will,  when  multiplied  into  the  divisor,  produce  the 
dividend. 


68  FUNDAiyiENTAL   RULES. 

2.  Divide  12b^c^y^  by  AbC'y^.  QuoL,  d¥c. 

3.  Divide  8cdx  by  dx.  QuoL,  8c. 
4  Divide  lOa'^cx^y  by  Ba^x^.  QuoL,  2acy. 

5.  Divide  lSa%^x  by  —  ^a'bx.  QuoL,  —  Qab. 

6.  Divide  —  20x^y*  by  —  Bx'^yK  QuoL,  4:xy-. 

7.  Divide  —  ^2a^m^-y^  by  la-m^.  QuoL,  —  Qay"". 

8.  Divide  —  Sla'bx'^y  by  da'by.  QuoL,  —  dax\ 

9.  Divide  —  l^x'^y-  by  —  13xy\  QuoL,  x. 

10.  Divide  Sa-c^  by  —  Sa^c^.  QuoL,  —  1. 

11.  Divide  Sa^m  by  da"m.  QuoL,  1. 

12.  Divide  —  ^^^/^  by  a;  "  ^?/  "  ^  QuoL,  —  xY\ 

13.  Divide  x"*  by  a;".  C^wo^.,  ^"^  -  ". 

14.  Divide  5a  "  W  by  2a6.  ^mo^.,  — . 

15.  Divide  Gah^  by  3  A^.  QuoL,  2a  "  ^6 

16.  Divide  lla~h~^x'^  by  11a " '6 -'j:^. 

17.  Divide  a^Z^'c" '  by  ab^c^  QuoL,  a'-'b'-'c-'^'+'l 


112*  I^rob,  2.  To  divide  a  jjolynomial  by  a  monomial. 
B  ULE. — DrviDE  each  teem  of  the  polynomial  dividend  by 

THE  MONOMIAL    DIVISOR,  AND  WRITE    THE    RESULTS   IN    CONNECTION 
WITH  THEIR  OWN  SIGNS. 

Dem. — This  rule  is  simply  an  application  of  the  corollaries  {103 f 
104:) ;  since  to  divide  a  poljTiomial  is  to  find  the  quotient  of  the  sum 
or  diiference  of  several  quantities,  which  by  these  corollaries  is  shown 
to  be  the  sum  or  difference  of  the  quotients  of  the  parts,     q.  e.  d. 


DIVISION.  69 

EXAMPLES. 

1.  Divide  l^a^x^  +  24^2^35  —  Vla^xc  by  Zax\ 

MODEL   SOLUTION. 

opEEATioN.  ^ax^)l^a?x'^  +  24«2^3^  —  Vla^xc 
5a-x~  ^  +  y^  —  4a^^  ~  *c* 
(  Explanation. — I  write  the  divisor  on  the  left  of  the  dividend,  and 
the  quotient  underneath  as  a  convenient  form.  Considering  the  first 
two  terms,  the  quotient  of  their  sum  is  the  sum  of  their  quotients 
{103)',  hence  I  divide  each  separately  and  add  the  results,  obtaining 
5a^x~^  -f- ^^^'  Again,  the  quotient  of  the  difference  of  these  two 
terms  and  the  third  is  the  difference  of  the  quotients  (104:)',  hence  I 
subtract  from  the  quotient  of  the  first  two  the  quotient  arising  from 
dividing  12a^xc,  which  is  4a*a;~^c,  and  have  for  the  entire  quotient 
5a^x-'  +  8a6  — 4a4a;-2c. 

2.  Divide  da^b'-  —  18a'¥  +  Ga^'b  by  Sab. 

QuoL,  a-b  —  6a^b  +  2a. 

3.  Divide  24:X'iy^  __  8^:4^5  —  24j??/2  by  8^7. 

QuoL,  dxy-  —  x^y^  —  3y\ 

4.  Divide  21a^x^  —  la^x^  +  14a^  by  lax. 

QuoL,  Sa-x-^  —  ax  -{-  2. 

5.  Divide  42as  —  lla'^  +  28a  by  la. 

QuoL,  6a2  _  JJ-a -f  4. 

6.  Divide  dk^'  —  2ik^c  +  48^-^  by  3^^. 

QuoL,  3^'o  —  8^'c  +  Wk. 

1.  Divide  12a^c^  —  48a'c4  —  S2a^c^  by  Wa^cK 

QuoL,  fa^c"^  —  Sa'G  —  2a'^c-\ 


8.  Divide  36m^  —  48m*  by  4m^. 


QuoL,  drn^—  Uvi^. 


3  3     2  1  1  7       2 

9.  Divide  m*  —  m^n^  by  m*.  QuoL,  wJ  —  mJ'^n^. 

10.  Divide  a^  —  a^b''  +  a^  by  A      QuoL,  a^  —  b^  +  a^. 


70  FUNDAMENTAL  RULES. 

11.  Divide  lla^  —  33a^  by  Hal  QaoL,  a^  -^  Sa"^. 

12.  Divide  72m  ^^  —  60m^/i^  by  24m'^. 

QuoL,   3m^  —  f/i^. 

13.  Divide  2a*  +  4ta^b  +  2a26^  by  2a''. 

QuoL,  a-'  +  2ab  +  62. 

14.  Divide  Ua'^x^  —  dab  ■{-  ISx^  by  3aa7. 

QuoL,  ^ax'^  —  r  +  67. 

15.  Divide  «'"  +  '  —  a'»  +  =^  —  a'"^^  —  «'"  +  •'  by  a'. 

g^o^.,  a'"-'  —  a---'  —  a'"  —  a"*  +  \ 


16.  Divide  5^"—  10^-"  +  l^x"y  by  527^ 


1  —2m 


QuoL,  X    "    —  2.^-^'"  ^■''^  -f  3.r    "    y. 

113*  Def. — A  polynomial  is  said  to  be  arranged  wth  reference  to 
a  certain  letter  when  the  term  containing  the  highest  exponent  of  that 
letter  is  placed  first  at  the  left  or  right,  the  term  containing  the  next 
highest  exponent  next,  etc.,  etc. 

L:.L.— The  polynomial  Q>x^y^  -\-  ^xy^  -f-  ix'-^y  +  y'  +  «*,  when 
arranged  according  to  the  descending  powers  of  y,  becomes  y*  -j"  ^^2/^ 
-}-  6x^y^  +  "Lc^y  -{-  x*.  In  this  form  it  also  clmnces  to  be  s>-frauged 
with  reference  to  the  ascending  powers  of  x. 


114:,  I^voh*  3*  To   'perform,   dividon    when  both   diu^" 
dend  and  divisor  are  polynomials. 

R  ULE. — Having  arranged  dividend  and  divisor  with  ee' 

TERENCE  TO  THE  SAME  LETTER,  DmDE  THE  FIRST  TERM  OF  THB 
DIVIDEND  BY  THE  FIRST  TERM  OF  THE  DIVISOR  FOR  THE  FIRST  TERM 
OF  THE  QUOTIENT.  ThEN  SUBTRACT  FROM  THE  DIVIDEND  THE  PRO- 
DUCT OF  THE  DIVISOR  INTO  THIS  TERM  OF  THE  QUOTIENT,  AND 
BRING  DOWN  AS  MANY  TERMS  TO  THE  REMAINDER  AS  MAY  BE  NECES- 
SARY TO  FORM  A  NEW  DIVIDEND.  DiVIDE  AS  BEFORE,  AND  CON- 
TINUE THE  PROCESS  TILL  THE  WORK  IS  COMPLETE. 


DIVISION.  71 

The  demonstration  of    this  rule  will  be  more  readily 
comprehended  after  the  solution  of  an  example. 

Ex. — Divide  6a'^x^  +  x*  —  4a^3  _|_  at  — ia^x  by  a;^  -f-  a'^ 
—2ax,  /  ^' 

MODEL  SOLUTION. 
DIVISOR.  DrVTDENa).  QUOTIENT, 

a^  —  2ax  -f  a;2)a-«  —  'ia^x  +  Ga'^x"^  —  4a^s  +  ^^(a^  —  2a*"  +  x^ 
a*  —  2a^x  +    a-x^ 

—  2a:-^x  +  5a-x'^  —  4:ax^ 


2a^x  +  4a2^2  —  2ax3 

a-x-^  —  2ax^  +  x* 
a^x^  —  2ax^  +  x^ 


Having  arranged  the  dividend  and  divisor  with  reference  to  the 
descending  powers  of  a,  and  placed  the  divisor  on  the  left  of  the  divi- 
dend, as  a  matter  of  custom,  I  know  that  the  highest  power  of  a  in 
the  dividend  is  produced  by  multiplying  the  highest  power  of  a  in  the 
divisor  by  the  highest  power  in  the  quotient.  Therefore,  if  I  divide 
a*,  the  first  term  of  the  arranged  dividend,  by  a^ ,  the  first  term  in 
the  arranged  divisor,  I  get  a^  as  the  highest  power  of  a  in  the 
quotient.  Now,  as  I  want  to  find  how  many  times  a^  —  2ax  -f-  *^i^ 
contained  in  the  dividend,  and  have  found  it  contained  a^  times  (and 
more),  I  can  take  this  a^  times  the  divisor  out  of  the  dividend,  and 
then  proceed  to  find  how  many  times  the  divisor  is  contained  in  what 
is  left  of  the  dividend.  Hence  I  multiply  the  di^dsor  by  a'^  and  sub- 
tract it  from  the  dividend,  leaving  —  2a^x  -f-  5a'x^  —  4ax^  +  x*'.  The 
same  course  of  reasoning  can  be  applied  to  this  part.  Thus  I  know 
that  the  next  highest  power  of  a  in  the  quotient  will  result  from  divid- 
ing the  first  term  of  this  remainder  by  the  first  term  of  the  divisor, 
etc.,  etc.  When  this  process  has  terminated  I  have  taken  a^,  and 
—  2ax,  and  x*  times  the  divisor  out  of  the  dividend,  and  finding 
nothing  remaining,  I  know  that  the  dividend  contains  the  divisor 
just  a*  —  2ax  +  ^^  times. 

We  will  now  give  tho  demonstration  of  the  rule. 

Dem.  — The  arrangement  of  dividend  and  divisor  according  to  the 
same  letter  enables  us  to  find  the  term  in  the  quotient  containing  the 
highest  (or  lowest  if  we  put  the  lowest  power  of  the  letter  first  in  our 
arrangement)  power  of  the  same  letter,  and  so  on  for  each  succeeding 
term. 


72  FUNDAMENTAL   RULES. 

The  other  steps  of  the  process  are  founded  on  the  principle,  thai 
the  product  of  the  divisor  into  the  several  parts  of  the  quotient  is 
equal  to  the  dividend.  Now  by  the  operation,  the  product  of  the  divi- 
sor into  the  first  term  of  the  quotient  is  subtracted  from  the  dividend  ; 
then  the  product  of  the  divisor  into  the  second  term  of  the  quotient  ; 
and  so  on,  till  the  product  of  the  divisor  into  each  term  of  the 
quotient,  that  is,  the  product  of  the  divisor  into  the  whole  quotient,  is 
taken  from  the  dividend.  If  there  is  no  remainder,  it  is  evident  that 
this  product  is  equal  to  the  dividend.  If  there  is  a  remainder,  the 
product  of  the  divisor  and  quotient  is  equal  to  the  whole  of  the  divi- 
dend except  the  remainder.  And  this  remainder  is  not  included  in  the 
parts  subtracted  from  the  dividend,  by  operating  according  to  the 
rule. 

EXAMPLES. 

[Note. — The  following  examples  are  arranged  for  division.  Let  the 
pupil  explain  his  solutions  according  to  the  model  preceding  the 
demonstration.  ] 

1.  Divide  x^  —  3ax-  +  Sa'^x  —  a^  hy  x  —  a. 

'   Quot,  x^  —  2ax  -f  a2. 

2.  Divide  2if  —  Idy^  +26?/  —  lQhjy  —  8. 

QuoL,  2y^  —  Sy-\-2. 

3.  Divide  x'^  —  a^x-^  +  2a^x  —  a*  by  x^  —  ax  +  a^. 

QuoL,  x'^  -\-  ax  —  aK 

4.  Divide  a^  +  45^  by  a^  —  2a6  +  2h\ 

OPERATION. 

<j-2  _  2a&  +  262)a4  +  ^h*  (a^  -f  2ah  +  262 

ai  —  2a^h  +  2a^h^ 

2a='6  —  2a^b^  +  W 
2a^h  —  4^262  -f  4aZ)3 

2«262  —  4a/>^  +  46» 
2a2&2  —  4^63  +  464 

5.  Divide  8a2  _  2(jah  +  1562  by  4a  —  36. 

QuoL,  2a  —  56. 

6.  Divide  a^  —  6-^  by  a  —  6.  QuoL,  a^  -\-  ah  -}-  b'K 


DIVISION.  73 

7.  Divide  a*  —  ia^x  -f  Qa-x-  —  4:ax^  +  x'^  by  a^  —  2ax 
+  x'K  QuoL,  a-  —  2ax  +  ^r^. 

8.  Divide  a^  +  x^  hy  a  -\-  x.  QuoL,  a-  —  ax  -}-  x:\ 

9.  Divide  48j73  _  TGa^^  —  Qla^x  +  lOoa^  by  2^;  —  3a. 

Quot,  24:X^  —  2ax  —  35a2. 

10.  Divide  2a-' +  a  —  Qhj  2a  —  3.  QuoL,  a  +  2. 

^     11.  Divide  j;^  _|.  7^  _j_  lo  by  j;  +  2.  ^wo^.,  a?  +  5. 

12.  Divide  x^  —  5x'  —  4:6x  —  40  by  j?  +  4. 

QuoL,  x-^  —  9^  —  10. 

13.  Divide  x^  —  dx^  +  27^  —  27  by  a;  —  3. 

QuoL,  x^  —  6^  +  9. 

[Note. — In  the  following  the  pupil  will  need  to  observe  whether  the 
terms  are  properly  arranged  or  not  before  commencing  the  division.  ] 

14.  Divide  5x-y  +  y^  -{-  x^  -\-  bxy-  by  ^xy  +  y^  +  x^. 

QuoL,  X  -\-  y. 

15.  Divide  6x'^y^  —  4:xy^  —  4:X^y  -}-  y^  -\-  x*  hj  x  —  y. 

QuoL,  x^  —  Sx^y  +  3xy'  —  y\ 

16.  Divide  Qx*  —  96  by  3^  —  6. 

QuoL,  2x^  +  4^2  +  8x  +  16. 

17.  Divide  3a^b^  —  3a'¥  —  ¥  -\-  a''  by  a^  —  b^  -\-  3ab^  — 3a^b. 

QuoL,  a^  +  3a-^6  +  Sat'  +  b^ 

18.  Divide  x'^  —  y^  —  Sx'^y  +  5xy*  -\-  lOx^y^  —  lOx^y^  by 
x  —  y.  QuoL,  (x  —  yy. 

19.  Divide  a'^  —  b^hj  a  —  b. 

QuoL,  a*  4-  a^b  +  a^b^  +  a¥  +  b^ 

/^  20.  Divide  tji-*  +  n-*  by  mi  +  n. 

21.  Divide  32^7^  +  243  by  2x  +  3. 

22.  Divide  b*  —  Zy^  by  b  —  y. 

QuoL,   63  +  b-^y  +  by^  +  y^  —  .^d!l_. 

b  —  y 


74  FUNDAMENTAL  EXILES. 

23.  Divide  a;^  -{-  px  -{-  q  hj  x  -\-  a. 

Quot,  X  -[■  p  —  a  -\- H —• 

X  -\-  a 

24.  Divide  a*  -\-  b*  by  a^  ^  abV2  +  t^ 

Quot. J  a'^  —  ab'^'A  +  b^. 

25.  Divide  ^x^  +  x^  +  i-x  +  ^  by  i^c  +1.     Quot,  x^  +  |. 

OPERATION. 

^X  +  1)1-^3  _|_  ^2  +    3^  +  f  (^^    +  f 

f  ^  +  f 

26.  Divide  a;*  +  2/^  ^y  ^  +  ?/• 


2v^ 

QWO^.,  a;3  5721/  -f  a7l/«  ?/3  + 


^+  y 

27.  Divide  a'""  +  2a"'6"  +  ^'"  by  a"'  +  b\       Quot.,  oT  +  6". 

OPERATION. 

28.  Divide  2a'"  —  6a'"6"  +  6a"6-"  —  26'"  by  a"  —  6". 

Qwo^.,  2a'"  —  4a"6"  +  26'". 

29.  Divide  ^x^  +  -if  j;^  _  a^  4.  |  by  ^x  +  3. 

(^140^.,  X'^  —  \x  +  |. 

30.  Divide  j:^  —  1/    by  ^^  —  ?/3.      Qmo^.,  x  +  ^"^j/^  -|   y. 

31.  Divide  a  —  6  by  a^  —  6*. 


3  11.         \  X        a 

Quot.,  a^  4-   a^6-^  4-  a-*6^   +  6* 

4r 


DIVISION.  75 

32.  Divide    x^  —  x^  —  ^x^  +  6j^  —  1x^  by  x^  —  4x^+  2. 

Quot.,  X  —  x'^. 

33.  Divide  a'"*  —  3a'"c"  +  2c'''  by  a"  —  c". 

^wo^.,  a"*  —  2(?". 

5      1  1 

34.  Divide  x^  4-  5x4 1 by  ^  H — . 

^         ^  x^ x-^     •^      ^  X 

QuoL,  a;*  4-  4  +  — . 

35.  Divide  x* by  x .       Quot.,  x^  4-  xA 1 . 

x^     -^  X         ^        '  X      x^ 

36.  Divide  -  —  2^2  _  ^^  +  f:^  by  2a  —  Zx. 

X  2         0-^ 

«""'••  2i  -  T- 

37.  Divide  ^  +  ^-^  +  f?by^+l 

2/3       12  ?/-^    '16    -^  3i/2      2y 

^    ^     3j:2        13^      39y 

38.  Divide  a^  +  («  _  i)j^2  _|_  (^  —  l):c3  +  (a  —  1)^^  —  x^ 
by  a  —  X. 

OPEEATION. 

a  —  x)a2  +  (a  —  l)a;2  +  (a  —  l)x3  -|-  (o  —  l)x'  —  x'>{a-\-x-\-x^-\-x^-\-xi 
a-  —  ax 


ax  -{-  [a  —  l)x^ 
ax  —  a;2 


ax-  +  («  —  l)ic^ 
ax-  —  x^ 


ax-^  -\-  (a  —  l)x* 
ax^  —  x^ 


ax^  —  x" 

ax^  —  x^ 


39.  Divide  x{x  —  l)a^  +  (^3  -^  2x  —  2)a^  +  (3^^  —  x^)a 
—  x^  by  fl-x  +  2a  —  x"^.  QuoL,   {x  —  l)a  +  x^ 


76 


FUNDAMENTAL  RULES. 


ScH.  —This  process  of  division  is  strictly  analogous  to  '  Long  Di- 
vision "  in  common  arithmetic.  The  arrangement  of  the  terms  corre- 
Bponds  to  the  regular  order  of  succession  of  the  thousands,  hundreds, 
tens,  units,  &c.,  while  the  other  processes  are  precisely  the  same  in 
both. 


Synopsis  for  Eeview. 


Division. 

Dividend. 

Divisor. 

Quotient. 

Remainder. 


Sch.  1.  Eelation  of  div'n  and  mult'n. 
Sch.  2.  General  reason  for  a  quotient. 
Cor.  1.  Multiplying  or  dividing  divi- 
dend and  divisor. 
Co7\  2.  Multiplying  or  dividing  div'd. 
Cor.  3.  "  "         "      divisor. 

Cor.  4.  Quotient  of  sum. 
Cor.  5.  "        "   difference. 


Def.  Cancellation. 


1  I   f  Law  of  Signs.     Dem. 

§  -b  I  ^  (  Cor.  1.  Meaning  of  exp't  0. 

i  i^  §^  [  Law  of  Expt's.    Dem.  •<  Cor.  2.  Negative  expt's. 

(  Cor.  3.  Transfer'ng  expt's. 


■i)   s2    ~ 


'1.  To    divide  one   monomial   by   another.      Rule. 
Dem. 

2.  To  divide  a  polynomial  by  a  monomial.      Rule. 

De7n. 

3.  To  divide  one  polynomial  by  another.     Rule. 

Dem. 
Scholium. 


Test  Questions. — How  do  negative  exponents  arise,  and  what  do 
they  signify  ?  What  is  the  value  of  any  quantity  with  0  for  its  expon- 
ent ?  Why  ?  How  do  you  divide,  when  dividend  and  divisor  consist 
of  the  same  quantity  affected  by  exponents  ?  Why  ?  Give  the  Gene- 
ral Rule  (2*ro&,  3)  and  its  demonstration. 


FUJS'DAMENTAL   PROPOSITIONS.  77 

CHAPTER  n. 

FACTORING. 


SECTION  I. 

Fundamental  Propositions. 

lis.  The  Factors  of  a  number  are  those  numbers 
which  multiphed  together  produce  it.  A  Factor  is,  there- 
fore, a  Divisor.  A  Factor  is  also  frequently  called  a  iniea- 
sure,  a  term  arising  in  Geometry. 

116.  A  Common  Divisor  is  a  common  integi-al 
factor  of  two  or  more  numbers.  The  Greatest  Common 
Divisor  of  two  or  more  numbers  is  the  greatest  common 
integral  factor,  or  the  product  of  all  the  common  integral 
factors.  Common  Measure  and  Common  Divisor  are  equiv- 
alent terms. 

117.  A  Com^non  MultiiJle  of  two  or  more  num- 
bers is  an  integral  number  which  contains  each  of  them  as 
a  factor,  or  which  is  divisible  by  each  of  them.  The  Least 
Common  Multiple  of  two  or  more  numbers  is  the  least  in- 
tegral number  which  is  divisible  by  each  of  them. 

118.  A  ComiJOsite  JS'uniher  is  one  which  is  com- 
posed of  integral  factors  difierent  from  itself  and  unity. 

119.  A  I* r hue  Number  is  one  which  has  no  in- 
tegral factor  other  than  itself  and  unity. 

120.  Numbers  are  said  to  be  Prime  to  each  other  when 
they  have  no  common  integral  factor  other  than  unity. 

ScH.  1. — The  above  definitions  and  distinctions  have  come  into  us« 
from  considering  Decimal  Numbers.  They  are  only  applicable  to  lit- 
eral numbers  in  an  accommodated  sense.  Thus,  in  the  general  "view 
which  the  literal  notation  requires,  all  numbers  are  composite  in  the 
sense  that  they  can  be  factored  ;  but  as  to  whether  the  factors  are 
greater  or  less  than  unity,  integral  or  fractional,  we  cannot  affirm. 


78  FACTORING. 

ScH.  2. — Skill  in  resolving  numbers  into  factors,  though  of  great 
importance,  can  only  be  attained  by  much  practice,  and  habits  of  close 
observation.  Yet,  if  the  learner  masters  the  following  propositions,  he 
will  lay  the  foundation  for  such  an  acquisition. 

12  H.  ^vop,  1,  A  monomial  may  he  resolved  into  literal 
factors  by  separating  its  letters  into  any  number  of  groups,  so 
that  the  sum  of  all  the  exponents  of  each  letter  shaWmake  the 
exponent  of  that  letter  in  the  given  monomiaL 


III.     5a%x^  may  be  resolved  into  5a  •  ah^  • 

11X1111 

a'^b^x^  X  a'^b'^x^  X  x^,  or  into  any  number  of  fac|ors,  in  a  like  manni^jv 

Dem. — This  is  a  direct  result  of  the  principle  that  monomials  are  mul- 
tipUed  by  writing  the  several  letters  in  connection,  and  affecting  esu^ 
with  an  exponent  equal  to  the  sum  of  the  exponents  of  that  letter  in 
the  factors. 

EXAMPLES. 

1.  Separate  12a"bx^  into  all  the  possible  factors  with  pos- 
itive integral  exponents. 

Factors,  3,  2,2,a,  a,  b,  x,  x,  and  x. 

3.     2 

2.  Separate  l^a^x^  into  two  equal  factors. 

_3_     1  _3_     1 

Factors,  4ai  ^x^,  and  4ai  ^x^. 

2. 

3.  Separate  %x^y^  into  two  factors,  one  of  which  is  4txy. 


4.  Bemove  the  factor  2{ax)'^  from  (Sa'^x. 


5.  Remove  the  factor  3a^  from  ISoc^. 


Factors,  4txy,  and  2x    ^y\ 

Result,  3a^A 

Result,  5^c-. 


6.  Resolve  m  into  two  factors. 

^  m,    ni^  •  771-%   771*    •  771*, 


I  1  2  13  1 

Results,  771^  •  771^",  v/  7n  -  v/  77)      fji^ .  771^,    771*    .  77^^, 


777^   -771^,  etc. 


FUNDAMENTAL   PROPOSITIONS.  79 

7.  Resolve  x  into  two  equal  factors.     Into  3.     Into  5. 


12^,  Prop,  2,  Any  factor  ichich  occurs  in  every  term 
of  a  jDolynomial  can  he  removed  by  dividing  each  term  of  the 
polynomial  by  it. 

Dem.  — This  is  the  ordinary  problem  of  division  by  a  monomials 
since  divisor  and  quotient  are  th©  factors  of  the  dividend. 


EXAMPLES. 

1.  Factor  3a  —  36. 

Result,  ^a  —  h). 

2.  Factor  ax  —  bx  -{-  ex. 

Result,  {a  —  b -\-  c)x. 

3.  Factor  5  —  5?/. 

Result,  5(1  —  y). 

4.  Factor  Qa'-y^  —  l%ay\ 

Result,  6ay^{ay  —  3). 

5.  Factor  42^^?/  —  14:xy  +  Ix^y^. 

Result,  7xy{Gx  —  2  -]-  x-y). 

6.  Factor  Slan-^x  —  6dcmx\ 

Residt,  9cma;(9m  —  Ix^). 

7.  Factor  la^x^y  —  21ax'^y\    Factoids,  lax^y  and  a^  —  ^y\ 

8.  Factor  llab'^x^  —  Mab^x'-  +  OGa^fc^a;^. 

Factors,  Vlab'^x^  and  6x  —  76  +  ^• 

a  Factor  924£i^b'^c^  —  lOlSa^b^c^  +  12S2a^b^c\ 

Factors,  lAa^b^c^  and  66a  —  lib  +  88c. 


123 •  JPtoJ),  3,  If  two  terms  of  a  trinomial  are  positive 
and  the  third  term  is  twice  the  jjroduct  of  the  square  roots  of 
these  two,  and  positive,  the  trinomial  is  the  square  of  the  sum 
of  these  square  roots.  If  the  third  term  is  negative,  the  tri- 
nomial is  the  square  of  the  difference  of  the  two  roots. 

Dem. — This  is  a  direct  consequence  of  the  theorems  that,  "The 
square  of  the  sum  of  two  quantities  is  the  sum  of  their  squares  plus 
twice  their  product  ;"  and  ' '  The  square  of  their  difference  is  the  sum 
of  their  squares  minus  twice  their  product. "     {94,  05), 


80  FACTORING. 

EXAMPLES. 

I.  "What  are  the  factors  of  a^  -^  ^ah  -[-  h-'i 

Model  Solution. — I  observe  that  the  two  terms  a^  and  h'^  of  thia 
trinomial  are  both  positive,  and  that  the  other  term,  2a6,  is  twice  the 
product  of  the  square  roots  of  a'^,  and  6^,  and  positive.     Hence  a  *  + 
'lah  -|-  &2  _  (■^  _|_  ^)2  —  (^a  -\-  h){a  -\-  h).     The  factors  are  a  +  6  and 
a  +  h. 

2.  What  are  the  factors  of  x'^  —  2ax  +  a^  ? 

3.  What  are  the  factors  of  m*  -\-  n^  -\-  2m-n'^  ? 

SuG. — Here  m*  and  n^  are  both  positive,  and  2m^n^  is  the  product 
t)f  their  square  roots.     Aris.   {m^  -\- n^)(m^  -\-n^). 

4.  What  are  the  factors  of  IGa'  —  8a  +  1? 

5.  What  are  the  factors  oi  m  -{-  2^y  mn  -{-  n '} 

Ans.,  (v/  7?i  +  v/ri)(\/  m  +  v^n). 

2  4  X     2 

6.  What  are  the  factors  of  ^^  +  ?/^  —  ^x'-^y^  1 

1  2  1  ^ 

Ans.,  x'-^  —  2/^  ^^<^  ^    —  V""' 

7.  What  are  the  factors  of  a"h  +  a^^  +  2a%^  ? 

8.  What  are  the  factors  of  x"^  +  2xy  —  y-^. 

Query. — Can  the  last  be  factored  according  to  this  Trob.  ?    Why  ? 

9.  If  4a'-',  1662,  and  IGab  are  the  terras  of  a  trinomial,  what 
must  be  their  respective  signs  so  that  the  trinomial  can 
be  factored  ? 

10.  Factor  I6a^b^m''  —  Sa^b'^m  +  1. 

Factors,  {ia-^b^m  — •  1)  and  {4:a%^m  —  1). 

II.  Factor  da-'  +  Sab  +  ^b^ 

Factors,  (3a  +  ^)  and  (3a  +  ^b). 

12.  Factor  49a^62  —  -^f-a6^  +  ib^. 

Factors,  {lab  —  p^)  and  {lab  —  %b'^]. 


V 


FUNDAMENTAL  PROPOSITIONS.  81 


13.  Factor  -^  +  2  +  — ' 


Factor 


14.  Factor  -^x^^  +  yV^/^  —  \oc^y'- 

Factors,  (fa;^  —  iy.)  and  {^x^  —  \y'). 

15.  Factor—  +  ^  —  2. 

?/->       x^ 

\y-        X-  /  \?/2         X   / 

16.  Factor  a^  ^  2av/^  +  ^. 

Factors,  {a  +  ^)  and  (a  +  v/j;). 

17.  Factor  j;  —  2b^  +  6-^. 

Factors,  {^x  —  h)  and  (v/a;  —  h). 

18.  Factor  ^  —  2Vx  ■  y  +  ?/. 

Factors,  {^x  —  v^y)  and  (v^^  —  ^y)- 

19.  Factor  a'h  —  "lax^h  +  x^. 

Factors,  {a^b  —  x)  and  {a^  —  x). 

20.  Factor  a  +  2v/a  +  1. 

Factors,  {\^a  +1)  and  (v^a  +  1). 


124,  J^vop,  4z.  The  difereiice  between  two  quantities  is 
equal  to  the  product  of  the  sum  and  difference  of  their  square 
roots. 

Dem. — This  is  an  immediate  consequence  of  the  theorem,  that  "The 
product  of  the  sum  and  difference  of  two  quantities  is  the  difference  of 
their  squares. "    Thus  a-  —  h-  =  {a-\-b){a  —  b).    Also,  if  we  have  m  —  n, 

m  is  the  square  of  ??i,  and  n,  of  n'    .'.  m  —  n  =^  {m^ -\- n^){m-  —  n^), 
etc. 

EXAMPLES. 

1.  Factor  16^'^  —  9y\ 

Factors,  {4:X  —  By)  and  (Ax  +  3y), 


82  FACTORING. 

2.  Factor  aHf  —  h'^xK 

Factors,  {ay  —  hx)  and  {ay  -\-  hx). 

3.  Factor  IQa^x'^  —  2562?/2. 

Factor Sy  {^ax  —  56y)  and  (4a^  +  56i/). 

4.  Factor  x^^  —  y*.        Factors,  {x^  —  ^/'O  ^^^  (^^  +  2/^)- 

2  2 

5.  Factor  x^  —  y^;  also  x^  —  if  ',    also  a;^  —  y'"^  ;  also 

j;  -4  —  y-*)  also  4a~^  —  96~\ 

11  1.        1. 

Factors   of  last  three,    {x'^  — ij^)  and  {x^  +  y^)   ; 

{x  -2  —  y--^)  and  {x-^  -{-  y -^)  \  and  {2a-^  —  W-^) 
and{2a-^  -\-3b-^). 

6.  Factor  m  —  n,  _ 

Factors,  {^  m  —  Vn)  and  (v^m  +  ^n). 

7.  Factor  2a  —  462. 

Factors,    2(v/  a  —  ^^26) (^  a  +  ^^26). 

a  Factor  25^"  —  3?y^". 

ni  m  

Factors,  {bx'^  —  ^'6if){^x~^ -^  ^'^y''). 


125.  I*rop,  o.  When  one  of  the  factors  of  a  quantity 
is  given,  to  find  the  other,  divide  the  given  quantity  by  the  given 
factor,  and  the  quotient  loill  be  the  other. 

Dem. — This  is  the  ordinary  problem  of  division,  since  the  divisor  and 
quotient  are  the  factors  of  the  dividend.  Any  problem  in  division  af- 
fords an  example. 
'  EXAMPLES. 

1.  Resolve  2a^  —  SQa^b  into  two  factors  one  of  which  is  2aK 

13  19 

2.  Remove  the  factor  — from — . 

x-         ?/3  x^         y^ 

o    -r»  ^y     e    j^       a^        3c    ^  .  9(?2m-2 

3.  Remove  the  factor  -r-  +  t: —  from  a^b  -* 


6     '     2m  4 

9       12         4 

4.  Remove  the  factor  3a  — -  +  2^  -•  from \- \-  — ;. 

a*      a'^x       x-^ 


FUNDAMENTAL   PROPOSITIONS.  83 

126»  I*TOp,  6»  The  difference  between  any  two  quanti- 
ties is  a  divisor  of  the  duterence  between  the  same  poivers  of 
the  quantities. 

The  SUM  of  two  quantities  is  a  divisor  of  the  diffeeence  of 
the  same  even  powers,  and  the  sum  of  the  same  odd  poivers  of 
th-e  quantities. 

Dem, — Let  X  and  y  beany  two  quantities  and  n  any  positive  integer. 
First,  X  —  y  divides  x"  —  y"  .  Second,  if  n  is  even,  x  -{-  y  divides 
K" — y*.    Third,  if  n  is  odd,  x  -^  y  divides  x"  -f-  V"- 

riKST. 

Taking  the  first  case,  we  proceed  in  form  with  the  division, 
till  four 

of    the    X — t/)x"  —  y"        (^x"  - '^ -\- x"  -  "y -\- x"  -  ^y-  -\- x"  -  ^y^ -\- &g. 
terms  of  x"  —  x ""  'y  ~         "^~~" 

the  quotient  x"  ~" '?/  —  y" 

(enough  to  deter-        x"  -  ^y  —  x"  -  "^y^ 


mine      the    law)  x"  ~  -y^  —  y" 

are  found.      We  find  x"  ~  ^y-  —  x"  -  ^y^ 

that   each    remainder  '     x" ""  y  —  j/" 

consists  of  two  terms,  the  second  of  x"  -  ^y^  —  x"  —  ^y* 

which,  —  y",  is  the  second  term  of  x"  ~  ^y^  —  y" 

the  di^'idend  constantly  brought  down 

unchanged  ;  and  the  first  contains  x  with  an  exponent  decreasing  by 

unity  in  each  successive  remainder,  and  y  with  an  exponent  increasing 

at  the  same  rate  that  the  exponent  of  x  decreases.     At  this  rate  the 

exponent  of  x  in  the  nth  remainder  becomes  0,   and  that  of  y,  n. 

Hence  the  ?ith  remainder  is  y"  —  y"  or  0  :  and  the  division  is  exact. 

SECOND    AND    THIED. 

X  -|-  j/)x"  4-2/"        ('x"  - 1  —  X"  -  ^v  -j-  X"  -  Hj'  —  X"  -  *y^  ,  &c. 
X"  -\-  x"-  ^y 

—  x"-i?/  +  2/" 

—  X"  -  hi  —  X"  -  "^y^ 

X"  -  h)^  ±  y^ 

Taking  x  -\-  y         x"  -  -y"^  -f-  x"  -  'y^ 

for  a  divisor,  we  —  x"  —  '^y^  +  y"^ 

observe  that  the  exponent        —  x"  -  ^y^  —  x»  —  *y* 
of  X  in  the  successive  re-  x'^  —  *y*  -^  y" 

mainders  decreases,  and  that  of  y  in- 
creases the  same  as  before.     But  now  we  observe  that  the  fii'st  term  of 


84  FACTORING. 

the  remainder  is  —  in  the  odd  remainders,  as  the  1st,  3rd,  5th, 
etc.,  and  -j-  i^i  the  even  ones,  as  the  2nd,  4th,  Gth,  etc.  Hence  if  n  is 
even,  and  the  second  term  of  the  dividend  is  —  y",  the  nth  remainder  is 
y*  —  2/"  ^^  ^'  ^^^  the  division  is  exact.  Again,  if  n  is  odd,  and  the 
second  term  of  the  dividend  is  -j-  y'\  the  nth  remainder  is  —  2/"  4~  !/" 
or  0,  and  the  division  is  exact,     q.  e.  d. 

'  127 »  ScH. — The  pupil  should  notice  carefully  the  form  of  the  quo- 
tient in  each  of  the  above  cases,  and  be  able  to  write  it  without  divid- 
ing. He  should  also  be  able  to  tell  at  a  glance,  whether  such  a  divi- 
sion is  exact  or  not,  and  if  not  exact,  what  the  remainder,  or  fractional 
term  of  the  quotient  is. 

EXAMPLES. 

Write  the  quotients  in  the  following  cases : 

1.  (^5  _  y5)  ^(x  —  y). 

QuoL,  X*  -f  x^y  +  x'iy^  -f  xy^  -f-  y*. 

2.  (a3  4-  63)   -^  (a  +  h).  QuoL,  a^  -—  a6  -f  b^ 

3.  (a3  _  53)  -^  (a  _  b). 

4.  {a*  —  m")  -^  (a  -{-  m). 

5.  {a^  —  771")  -i-  (a  —  m), 

6.  (m7  -f  ?i7)  -^  (tti  +  n). 

7.  (m8  —  6«)  ^  (m  -f  b),  also  by  (m  —  b). 

SuG. — Notice  that  x^  ^  is  the  4th  (an  even)  power  of  x^,  and  y''  *  is  the 
same  power  of  y^.  It  may  be  best  for  the  pupil  to  obtain  this  quotient 
thus.  Put  x^  =  a,  and  y^  =  b.  Then  x'  =^  =  a"*  and  y^^  =  b'^.  But 
(a^  —  b'^)  ~  (a-}-6)  =  a^  —  a^b  -{-  ab'^  —  b\  Whence  restoring  the 
valaes  of  a  and  b,  we  have  (x^^  —  y^^)  -r-  {^^  -h  V^)  =  x^  —  x^y'^  -f- 
^^y^  —  2/^'  since  a^  =  x^,  — a^b=  —  x^y'^,  -\-ab'^  =  +  '^^y^,  and  — 

9.  (tti'o  -f  ni«)  ~  (m2  -f  n^). 

Quot,  mP  —  m^n^  -f  m^w^  —  m^^  -f-  n^. 

10.  (a;'8  —  ?/'8)  -:-  {x^  —  y^),  also  by  {x^  +  y^). 

11.  (a%i«  —  2/^)  -^  (a?n'  —  y). 


FUNDAMENTAL   rEOPOSITIONS.  85 

Sua— Notice  that  a^m^  is  the  third  power  of  am^,  as  y^  is  of  y.  The 
quotient  mav  be  obtained  by  first  using  a  single  letter  for  «m^,  as  x, 
\\Titing  out  [x^  —y')^iz  —  y)==  x^  -^xy  ^y-,  and  finally  restor- 
ing the  value  of  a*.  After  a  Uttle  practice  the  pupil  will  not  need 
to°go  through  such  a  process,  but  can  write  the  quotient  at  once. 
This  quotient  is  a'^m^  -\-  am'^y  -f-  y^- 

12.  (1  -  7/)  -^  (1  -  y). 

.    13.  {ye  —  l)^{y  +  1),  also  by  (y— 1). 

14  {a'x^  +  5io?/'o)  -i-  {ax  +  b^y"^). 

15.  {16x*  —  8Uf)  ^  {'2x  4-  3y2),  also  by  {2x  —  S?/^). 

[Note.— In  the  following  examples  be  careful  to  observe  in  which 
cases  the  division  is  exact  ;  but  if  it  is  not  exact,  the  quotient  should 
be  WTitten  with  the  fractional  term.  ] 

16.  Is  a^  +  b\  exactly  divisible  hj  a  -\-  b,  or  a  —  b? 

17.  Is  m6  — y^,  exactly  divisible  by  m  +  y,  or  m  —  ?/  ? 

18.  Write  the  following  quotients  {x"'"'—y"")  -^  {x  -\-  y), 
also  hj  X  —  y.  (a;-™  +  ^  -^  y"'"^'^)  -^  {^c  -\-  y),m  being  an  in- 
teger. 

QuEET. — Is  the  division  exact  in  both  cases  in  the  last  example  ?  Is 
2m  even  or  odd  ?    Is  2m  -}-  1  even  or  odd? 


FOR  REVIEW  OR  ADVANCED  COURSE. 

[Note. — The  following  corollary  and  examples  may  be  omitted  till 
after  the  pupil  has  been  through  the  chapter  on  Powers  and  Boots,  if 
thought  desirable.  ] 

128,  CoR. — The  last  proposition  applies  equally  to  cases 

I  JL 

involving fractional  or  negative  exponents.     Thus,  x^  —  y^ 

A.  4. 

divides  x^  —  y^,  since  the  latter  is  the  difference  between 
the  4th  powers  of  x'^  and  y'^.     So  in  general  x~^  —  y~~ 

an  ax 

divides  j;~™ — y~~,  (^  being   any  positive   integer.     This 

n  s 

becomes   evident   by    putting   x~m  =  v,  and  y~T  =^  w; 


86  FACTORING. 

an  an 

whence  x  ~  ^  =  v",  and  y~  T  =  nf.     But  i/*  —  lu"  is  divisi- 

an  as  n  s 

ble  by  v — lu,  hence  x  ~  ^  —  y~T  i^  divisible  hj  x~~^  —  y    ^. 

EXAMPLES. 

19.  What  is  the  quotient  of  (a"'  —  &"')  -^  (a~'  —  &-')? 

20.  What  is  the  quotient  of  (^~^— y""^)  -^  (^~^+  2/"^)? 

^ns.,  ^     ^  —  X     ^y    ^  -\-  X    ^y     ^  —  y    *. 

21.  What  is  the  quotient  of  [^  —  if~\  -^  f-  4  2/'  1  ? 

—  X  — JL  _i 

22.  Is  ;r    '^ — ?/ ~" '' divisible  by  a;    ^ — y~"^or^     ^  +  2/~^? 

23.  Is  ^73  —  2/3  divisible  by  ^x  -\-  ^y  or  "^x  —  \^y  ? 

24.  Is  ^3  _[-  2/3  divisible  by  ^^x  -\-  ^y  or  ^x  —  ^y  ? 

25.  What  is  the  quotient  of  (ax^  +  h\j)^{a^  ^x  +  V^?/^)? 


120,  Pvo])»  7.  -4  trinomial  can  be  resolved  into  iico  bi- 
nomial factors,  when  one  of  its  terms  is  the  product  of  the 
tquare  root  of  one  of  the  other  two,  into  the  sum  of  the  factors 
of  the  remaining  term.  The  two  factors  are  severally  the  alge- 
braic sum  of  this  square  root,  and  each  of  the  factors  of  the 
third  term. 

III. — Thus,  in  a;'  -f-  '^''^  +  10,  we  notice  that  7a;  is  the  product  of  the 
square  root  of  x* ,  and  2  -f-  ^  (tiie  sum  of  the  factors  of  10).  The  factors 
of  x^  -\-1x  -\-  10  are  X  -f-  2  and  x  -\-^.  Again  x^  —  3x  —  10,  has  for 
its  factors  x  -f-  2  and  x  —  5,  —  3x  being  the  product  of  the  square 
root  of  x^  (or  x),  and  the  sum  of  —  5  and  2,  (or  —  3),  which  are  factors jof 
^  10.  Still  again,  x^  -f  3x  —  10  =  (x  —  2)(x  -f-  5),  determined  in  the 
same  manner. 

Dem. — The  truth  of  this  proposition  appears  from  considering  the 
product  of  X  -|-  ^  ^y  ^  +  ^'  which  is  x^  +  (a  -f-  &)  *  +  «&•  ^  this 
product,  considered  as  a  trinomial,  we  notice  that  the  term  (a  -\~  h)  x,  is 


FUNDAMENTAL   niOPOSITIONS.  87 

the  i)roduoi  of  ^  x*  and  a  +  ?>,  the  sum  of  the  factors  of  ah.  lu  hke 
manner  (cc  -f-  o.){x  —b):=x'^  -j-  (a  —  b)x  —  a&,  and  (x  —  a){x  —  b) 
=:  x^  —  {a  -{-b)x  -}-  «^>  both  of  which  results  correspond  to  the  enun- 
ciation.   Q.  E.  D. 

[Note. — In  application,  this  proposition  requires  the  solution  of  the 
problem  :  Given  the  sum  and  product  of  two  numbers  to  find  the 
numbers,  the  complete  solution  of  which  cannot  be  given  at  this  stage 
of  the  pupil's  progress.  It  will  be  best  for  him  to  rely,  at  present,  sim- 
ply upon  inspection,  ] 

EXAMPLES. 

1.  Show  that,  according  to  this  proposition,  the  factors 

of  ^2  _|_  3^  _|_  2  are  x  -{-  1  and  ^  +  2.  Verify  it  by  actual 
multiplication. 

2.  In  hke  mannei'  show  the  following  : 

a;2_  7^_^   12    ^    (j;_3)(^_4). 

x^  — j;— 12    =  (j7  + 3)(j;  — 4). 
^2  4_^__12    =^  (^_3)(^  +  4). 

v^  —  5^  +  6     — - 
a;2  +  207  —  35   =^ 


130,  I^vop,  S.  W(y  can  oft  m  detect  a  factor  by  separat- 
ing a  polynomial  into  parts. 

Dem. — The  correctness  of  this  process  depends  upon  the  principle 
that  whatever  divides  the  parts,  divides  the  whole. 

EXAMPLES. 

1.  Factor  x^  +  12a?  ^  28. 

Model  Solution. — ^The  form  of  this  polynomial,  suggests  that  there 
may  be  a  binomial  factor  in  it,  or  in  a  part  of  it.  Now  x'~  —  -Lx-j-'L  is 
the  square  of  x  —  2,  and  {x^  —  4ic  -(-  4)  +  (16x  —  32)  makes  x'^  + 
12x  —  28.  But  {x^  _  4.C  +  4)  4-  (16x  —  32)  =  (x  —  2)(x  —  2)  + 
(X  —  2)16  =  (X  —  2)(x  —  2  -f  16)  =  (X  —  2)(x  +  14).  Whence 
X  —  2,  and  x  -^  14:  are  seen  to  be  the  factors  of  x*  +  12x  —  28. 


88  FACTORING. 

2.  Resolve  12a-b  +  3&?/-  —  15aby  into  its  factors. 

Solution.  3b  is  seen  to  be  a  common  factor,  and  removing  it,  we 
have  4a^  -\-  y'^  —  5f^y-  This  latter  polynomial  may  be  separated  into 
the  parts  a*  —  2ay  -f-  y^,  and  3a^  —  Say.  Hence,  4a^  -\-  y^  —  Say 
=  {a'  _  2ay  +  2/2)  +  (Sa^  —  3ay^  =  (a  —  y)(a  -  ij)  +  («— t/)3rt  = 
(^a —  y){a  —  y  -{-  3a)  =  (a  —  2/)(4a  —  y).  Hence,  the  factors  of 
X2a^b  +  3by^  —  15aby,  are  36,  a  —  y,  and  4a  —  y. 

3.  Factor  Ga-^  +  15a^6  —  4:a^c''  —  10a-bc\ 

Solution,  a-  is  evidently  a  factor.  Bemoving  this,  we  have  6a^  -f" 
15a-6  —  4ac-  —  106c'-.  The  latter  expression  may  be  written  3a-(2a  -f-  S^) 
—  2c^(2a  +  56)  =  (3a'2  — 2c-2)(2a  +  56).  Henoe  6a'^-{-15a^h  —  4.a^c^  — 
10a26c2  =  a2(3c»2  —  2c-2)(2a  +  56). 

4.  Factor  Sa'^  —  2a  —  1.     Factors,  a  —  1,  and  3a  +  1. 

5.  Factor  5a^  —  Sax  +  'Sx'\     Factors,  5a  —  3x  and  a  —  x. 

6.  Factor  20ax  —  2ban  —  \(Sbx  -f  2Qbn. 

Factors,  4:X  —  5/1  and  5a  —  46. 

7.  Factor  Sa^  +  22ab  +  1562. 

Factors,  2a  +  36,  and  4a  +  56. 

8.  Factor  \2x^  —  Sxy  —  ^x^y^  +  &y^. 

Factors,  3ar2  —  2i/,  and  4a;  —  3?/2. 


SECTION  11. 


y  Greatest  or  Highest  Common  Divisor. 

131,  Def. — It  is  scarcely  proper  to  apply  the  term  Greatest  Com- 
mon Divisor  to  literal  quantities,  for  the  values  of  the  letters  not  being 
fixed,  or  specific,  great  or  small  cannot  be  affirmed  of  them.  Thus, 
whether  a^  is  gi'eater  than  a,  depends  upon  whether  a  is  greater  or  less 
than  1,  to  say  nothing  of  its  character  as  positive  or  negative .  So,  also, 
we  cannot  with  propriety  call  «•'  —  y^  greater  than  a  —  y.  If  a  =  |-, 
And  y  =  \,  a'^  — y'^  =  ^^,  and  a  —  ?/  =  i 5  .* .  in  this  case  a^  —  2/"^<C  ^ 
. —  y.  Again,  if  a  and  y  are  both  greater  than  1,  but  a  <C  y,  a-'  —  y'^ 
jthough  yinmerically  greater  than  a  —  y  is  absolutely  less,  since  it  is  a 
greater  negative. 


\j' 


GREATEST   OR   HIGHEST   C.    D.  89 

Instead  of  speaking  of  G.  C.  D.  in  case  of  literal  quantities,  we  should 
speak  of  the  Highest  Common  Divisor,  since  what  is  meant  is  the  divi- 
sor which  is  of  the  highest  degree  with  reference  to  the  letter  of  arrange- 
ment, i.  e.,  involves  the  highest  power  of  that  letter. 

[Note. — The  general  rule  for  finding  the  Greatest  or  Highest  Com- 
mon Di-sisor  is  founded  upon  the  four  following  lemmas.] 

132.  Lemma  1. — The  Greatest  or  Highest  G.  D.  of  two  or 
more  numbers  is  the  product  of  their  common  prime  factors. 

Dem. — Since  a  factor  and  a  divisor  are  the  same  thing,  all  the  com- 
mon factors  are  all  the  common  divisors.  And,  since  the  product  of 
any  number  of  factors  of  a  number  is  a  divisor  of  that  number,  the 
product  of  all  the  common  prime  factors  of  two  or  more  numbers  is  a 
common  ditisor  of  those  numbers.  Moreover,  this  product  is  the 
Greatest  or  Highest  C.  D.  since  no  other  factor  can  be  introduced  into 
it  without  preventing  its  measuring  (dividing),  at  least,  one  of  the 
given  numbers.     Q.  E.  D. 

EXAMPLES. 

1.  What  is  the  G.  C.  D.  of  48,  108,  and  72  ? 

Model  Solution.  48  ^  2  -  2  •  2  •  2  ■  3,  108  =  2  •  2  ■  3  •  3  •  3,  and  72  = 
2  •  2  -  2  •  3  -  3.  The  common  factors,  or  divisors,  are  2  •  2  •  3  ;  hence, 
2-  2-  3  =  12,  divides  all  the  numbers,  and  is  a  C.  D.  Moreover,  as 
there  is  no  other  common  factor,  if  we  introduce  an  additional  factor 
into  2  •  2  -  3,  it  \\ill  not  divide  the  given  number  or  numbers  which  do 
not  contain  this  factor.  Thus,  if  we  introduce  a  factor  5,  it  will  not 
divide  any  one  of  the  given  numbers.  If  we  introduce  another  factor, 
2,  and  have  2  •  2  •  3  •  2  =  24,  it  will  not  divide  108,  which  has  but  two 
factors  of  2.     .  •.  2  •  2  •  3  =  12,  is  the  G.  C.  D.  of  48,  108,  and  72. 

[Solve  the  following,  giving  the  explanation  as  above.  ] 

2.  Find  the  G.  C.  D.  of  84,  126,  and  210.     G.  G.  D.,  42. 

3.  Find  the  G.  C.  D.  of  70,  105,  and  245.     G.  G.  D.,  35. 

4.  Find  the  Highest  C.  D.  of  I'lab^c  and  1^h^€\ 

Solution. — Here  we  see  at  once  the  Highest  Common  Divisor  with 
reference  to  the  letters,  as  6,  h,  and  c  are  all  the  common  literal  fac- 
tors.    There  is  no  common  factor  in  12  and  25.     Hence  6^  c*  is  the  H, 

*  Whether  b^c.  is  the  greatest  C.  D.,  depends  upon  the  values  of  6  and  c.  If  eitheJ 
is  a  proper  fraction,  as  b,  and  c  an  integer,  c  is  greattr  than  b^c. 


90  FACTORING. 

CD.,  inasmuch  as  no  other  factor  can  be  introduced  into  this  pro- 
duct (b'^c)  and  it  still  remain  a  divisor  of  12ab^c  and  255 ■*c^. 

5.  Find  the  H.  C.  D.  of  Ua^b^c^-  and  Sa'^b^c. 

6.  Find  the  H.  C.  D.  of  Ga^^^  l^a?xy,  and  2ia^x-^y\ 

7.  Find  the  H.  C.  D.  of  ^x^y^  and  "Imnxy. 

H.  G.  D.,  XAf. 

8.  Find  the   H.  C.  D.  of  ^a-x  —  (Sahx  -f  Wx  and  ^a-y 
-AbHj. 

Suggestions,     ^a^x  —  Gahx  -f-  ^h'^x  =  3x{a  —  b){a  —  h);  and  4:a^y 

—  'ib'^y  =  4.yya  +  b){a  —  b).  .  •.  H.  C.  D.  is  a  —  6. 

9.  Find   the   H.   C.  D.  of  x'^  +  12a;  —  28  and  x^  +  ^x^ 

+  21x  —  98. 

Suggestions. — Factor  the  polynomial  of  the  lower  order  first,  as  it  is 
more  easily  resolved,  ic*  -|"  1^*  —  '^^'  ^^^J  ^^  factored  by  {120}  or 
{130).     According  to  the  latter  process  we  have  x'^  -\-  12x  —  28  =  x"^ 

—  4x  +  4  +  (16.C  —  32)  ^  (x  —  2)2  +  16',x  —  2)  =  (a;  —  2) 
(ic  —  2  -]-  16)=  {X  —  2)(x  -|-  14).  Now  find  by  trial  whether  either  or 
both  of  these  factors  are  divisors  of  the  other  polynomial,  x  —  2  is, 
and  X  -{-  14  is  not.     .-.x  —  2  is  the  H.  C.  D. 

10.  Find  the  H.  C.  D.  of  ^a'^x'^y  —  ^a'^xy  —  ^Qa'^y  and  ^a'^x^ 
—  48a-'^  —  3a-^^^  +  48a2.     H.  G.  D.,  Sa^o;  —  12a2. 

133.  ScH. — The  difficulty  of  factoring  renders  this  process  im- 
practicable in  many  cases.  There  is  a  more  general  method.  But,  in 
order  to  demonstrate  the  rule,  we  must  prove  three  additional  lemmas. 

[Note.— These  lemmas  and  the  demonfitration  oi  ih.&  general  rule  may 
be  omitted, if  the  teacher  thinks  best,  till  a  review.  But  the  rule  should 
be  thoroughly  learned  and  its  apphcation  rendered  perfectly  famihar, 
even  if  the  demonstration  is  omitted,  at  first.  The  author  would  not 
omit  it  at  all,  but  has  indicated  the  omission  in  deference  to  others.] 

1S4.  Lemma  2. — A  polynomial  of  the  form  Ax"  +  Bx"  ~  ^ 

+  Cx"~* Ex  +  F  ivhich  has  no  common  factor  in  every 

term,  has  no  dimwr  of  its  own  degree  except  itself. 


HIGHEST   C.   D.  91 

By  this  form  it  is  meant  that  all  Uke  powers  of  the  letter  of  arrange- 
ment are  collected  into  a  single  term,  and  that  the  polynomial  is  ar- 
ranged according  to  some  letter,  as  for  division. 

Dem. — 1st.  Such  a  polynomial  cannot  have  one  factor  of  the  nth 
degree, — its  own, — with  reference  to  the  letter  of  arrangement,  and 
another  which  contains  the  letter  of  arrangement,  for  the  product  ol 
two  such  factors  would  be  of  a  higher  (or  different)  degree  from  the 
given  poljTiomial. 

2d.  It  cannot  have  a  factor  of  the  nth  degree  with  reference  to  the 
letter  of  arrangement,  and  another  factor  which  does  not  contain  that 
letter,  for  this  last  factor  would  appear  as  a  common  factor  in  every 
term,  which  is  contrary  to  the  hypothesis,  q.  e.  d. 

III.  4m^  —  5ay  -|-  y^  cannot  have  a  factor  containing  a^,  such  as 
a2  —  y,  and  another  which  contains  a,  as  4:a  —  2y,  since  two  such  fac- 
tors multipUed  together,  would  give  a  higher  power  of  a  than  a^ ,  to  say 
nothing   of  other  terms.     Again,  it  cannot  have  such  a  factor  as  a* 

—  ay  —  y  and  another  which  does  not  contain  a,  as  4:  —  5y  -\-  y^, 
since  this  last  would  then  appear  as  a  factor  in  every  term  of  the  pro- 
duct when  arranged  -with  reference  to  a;  as  (4  —  5y  -\-  y^)a^  —  (4 

—  5?/  -j-  y^)ay  —  (4  —  ^y  -\-  y^)y.  But  the  hypothesis  is  that  the 
given  polynomial  shall  have  no  common  factor  in  every  term. 

13S.  Lemma  3. — A  divisoi^  of  any  number  is  a  divisor  cf 
any  midtijjle  of  that  number. 

III. — This  is  an  axiom.  If  a  goes  into  b,  q  times,  it  is  evident  that  it 
goes  into  n  times  h,  or  nb,  n  times  q,  or  nq  times. 

130,  Lemma  4. — A  common  divisor  of  two  numbers  is  a 
divisor  of  their  sum  and  also  of  their  difference. 

Dem. — Let  a  be  a  C.  D.  of  m  and  n,  going  into  m,  p  times,  and  into 
n,  q  times.     Then  {m  -{-  71)  -^  a  =  p  -\-  q.     q.  e.  d. 

III. — If  this  appears  a  Httle  abstract  and  unsatisfactory  to  the  learn- 
er, let  him  illustrate  it  thus :  4  is  contained  in  20,  5  times,  and  in 
8,  2  times  ;  hence  it  is  contained  in  20  +  8,  5  +  2,  or  7  times,  and  ia 
20  -  8,  5  -  2,  or  3  times. 


92  FACTORING. 

137,  Proh, — To  find  the  H.  C.  D.  of  tivo  j^olynovnals 
without  the  necessity  of  7'esolving  them  into  their  prime  fac- 
tors. 

BULK — 1st.  Arranging  THE  polynomials  with  reference 

TO  THE  SAME  LETTER,  AND  UNITING  INTO  SINGLE  TERMS  THE  LIKE 
POWERS  OF  THAT  LETTER,  REMOVE  ANY  COMMON  FACTOR  OR  FACTORS 
V/HICH  MAY  APPK\R  IN  ALL  THE  TERMS  OF  BOTH  POLYNOMIALS,  RE- 
SERVING THEM  AS  FACTORS  OF  THE  H.  C.  D. 

2nd.  Reject  from  each  polynomial  all  other  factors  which 

APPEAR  IN  each  TERM  OF  EITHER. 

3rd.  Taking  the  polynomials,  thus  reduced,  divide  the  one 

WITH  THE  GREATEST  EXPONENT  OF  THE  LETTER  OF  ARRANGEMENT, 
BY  THE  OTHER,  CONTINUING  THE  DIVISION  TILL  THE  EXPONENT  OF 
THE  LETTER  OF  ARRANGEMENT  IS  LESS  IN  THE  REMAINDER  THAN  IN 
THE   DIVISOR. 

4th.  Reject  any  factor  which  occurs  in  every  term  of 

THIS  remainder,  AND  DIVIDE  THE  DIVISOR  BY  THE  REMAINDER  AS 
thus  reduced,  TREATING  THE  REMAINDER  AND  LAST  DIVISOR  AS 
THE  FORMER  POLYNOMIALS  WERE.  CONTINUE  THIS  PROCESS  OF  RE- 
JECTING FACTORS  FROM  EACH  TERM  OF  THE  REMAINDER,  AND  DIVID- 
ING THE  LAST  DIVISOR  BY  THE  LAST  REMAINDER  TILL  NOTHING  RE- 
MAINS. 

If,  at  ANY  TIME,  A  FRACTION  WOULD  OCCUR  IN  THE  QUOTIENT, 
MULTIPLY  THE  DIVIDEND  BY  ANY  NUMBER  WHICH  WILL  AVOID  THE 
FRACTION. 

The  LAST  DIVISOR  MULTIPLIED  BY  ALL  THE  FIRST  RESERVED  COM- 
MON FACTORS  OF  THE  GIVEN  POLYNOMIALS,  WILL  BE  THE  H.  C.  D. 
SOUGHT. 

Dem. — We  will  first  give  a  demonstration  of  this  rule  by  means  of  a 
particular  example.  Let  it  be  required  to  find  the  H.  C.  D.  with  ref- 
erence to  a,  of  the  following  polynomials  : 

I'ia^i)^  +  3f>'^^2  _i5a62y  -f.  12a^by  -f  'Shy'  —  15ahy-,  and  Qa-'b^  — 
ea'b^'y  -  '^b^y'  -j-  2ab-y-  -j-  Ga'by  —  Qa^by^  —  ^by"  -f  2abyK 


HIGHEST  C.  D.  93 

OPEBATION. 

12a«6*  +  3b^y^  —  15ah'y  +  Vla-by  +  dby"^  —  15aby^ (A). 

Gaib^—6a^b^y—2b^y^  -\-2ab^y^  -^6a^by—Ga^by^~2by'^-{-2aby-'  -   (B). 

(C). 

(E). 

(F). 

(G)  (II) 

4a2  —  5ya  +  y^)Sa^  —  Sya^^  +  y-a  —  y^ 

■i 


ia-6  H-  by-  - 
3a^b  —  3a-' by 

-  oaJjy  -f-  4:a''y  -j-  y^  — 
—  by-^  +  aby^  +  Sa^; 

-  5ai/^      -  - 

y  —  ^a'y'- 

-y' 

■  +  «?/'  -  - 

(^■h'j  -j-  4.y,a-^  — 
(•%  +  3.VV/^  - 

{5by  -f-  oy-)a-]-  {by- 
(3by  +  3y')a^  +  (^v* 

+  r)    - 
+  ?/■■*)«  — 

■  ihy' 

+  y^)  --- 

(ij  -- 

(K)  -  . 

12a^  - 

. 12a3  - 

-  15?/a2  +  3y-'a 

(L)     - 

■  '^y^i''^  4-y'^  — V 

4 

(M)  - 
CN)  -  - 

(0)  -. 

■  -  Reject  l'Ji/2 

a  —  v)4a=^  — 

G 

i-f  ?/*(4a  — y. 
4^2  _  4j/tf 

-  2/«  +  y' 

—  ya-\-y^ 

.-.  The  H.  C.  D.  of  (A)  and  (B)  is  {b){b  +  y){a  —  y)  =  ab^   +  f% 

liEASONING. 

1st.  Removing  Z>  from  both  (A)  and  (B),  and  reserving  it  as  a  factor 
of  the  H.  C.  D.  (Lem.  1),  and  rejecting  3  from  (A),  and  2  from  (B), 
since  they  are  not  common  factors,  and  hence  cannot  enter  into  the 
H.  C.  D.,  we  get  (C)  and  (D). 

2nd.  Arranging  (C)  and  (D)  ^vdth  reference  to  a,  the  letter  with  re- 
spect to  which  the  H.  C.  D.  is  sought,  both  for  convenience  in  divid- 
ing, and  to  observe  if  any  other  common  factor  appears,  we  have  (E) 
and  (F).  In  these  we  readily  discover  and  remove  the  common  factor, 
(6  -\-y),  reserving  it  as  a  factor  of  the  H.  C.  D.,  and  get  (G)  and  (H)i 

3rd.  Having  now  found  two  of  the  common  factors  of  (A)  and  (B), 
and  removed  some  which  were  not  common,  it  remains  to  determine 
whether  there  are  any  more  common  factors,  that  is,  whether  there 
is  a  C.  D.  in  (G)  and  (H). 

(G)  is  its  own  H.  D.,  Lem.  2  ;  hence  if  it  divides  (H),  it  is  the  H. 
C.  D.  of  (G)  and  (H).     "We,  therefore,  try  it.     Dividing,  4a^  goes  into 


94  FACTORING. 

3a^,  fa  times  ;  but,  to  avoid  fractions,  we  multiply  (H)  by  4,  since,  as 
4  is  not  a  factor  of  (G),  the  H.  C.  D.  of  (G)  and  4  times  (H)  is  the 
same  as  of  (G)  and  (H).  We  thus  obtain  (I).  .  • .  K  (G)  divides  (I)  it 
is  the  H.  C.  D.  of  (G)  and  (H).  Trying  it,  we  find  the  remainder 
(L).  Now,  any  C.  D.  of  (G)  and  (I)  is  a  divisor  of  {K)[a  multiple  of 
(G)],  Lem.  3,  and  of  (L),  Lem.  4.  And  we  now  have  to  find  the  H.  C 
D.  of  (G)  and  (L),  upon  which  we  reason  just  as  upon  (G)  and  (H) 
Thus,  as  (G)  is  its  own  only  divisor  of  the  2nd  degree,  if  it  divides  (L) 
or  (M) [since  (M)  =  4  (L)  and  4  is  not  a  factor  in  (G)],  it  is  the  H.  C 
D.  of  (G)  and  (H).  Trying,  we  find  a  remainder  (O).  Now,  any  di- 
visor of  (G)  and  (M)  is  a  divisor  of  (N),  Lem.  3,  and  of  (0),  Lem.  4, 
The  question  is  therefore  reduced  to  finding  the  H.  C.  D.  of  (G)  and  (O) 
Upon  these  we  reason  as  before.  Hejecting  Idy^  since  it  is  not  a  fac 
tor  of  (G),  and  hence,  cannot  enter  into  the  H.  C.  D.,  we  have  (P) 
The  H.  C.  D.  of  (G)  and  (P)  cannot  be  higher  than  (P),  and  as  (P)  is 
its  own  only  divisor  of  its  own  degree,  if  it  divides  (G, )  it  is  the  H 
C.  D.  sought.     It  does.     .  •.    a  —  y  is  the  R.  G.  D.  of  (G)  and  (H). 

Finally,  h,  {h  -{-  y),  and  (a  —  y)  are  all  the  common  factors  of  (A)  and 
(B)  and  hence,  6  (&  +  y){'^  —  V)  =  «^^  +  «^2/  —  &*2/  —  ^V^  is  their 
H.  C.  D. 

ScH. — It  often  occurs  that  one  or  more  of  the  above  steps  are  not  re- 
quired, especially  the  removing  of  a  compound  factor  from  the  given 
polynomials. 

EXAMPLE. 

Find  the  H.  C.  D.  with  respect  to  x,  of  a:'  —  8x^  -}-  ^Ix" 
—  20^  +  4,  and  2x^  —  12^2  _J_  21^;  _  10. 

Model  Solution. — Calling  the  2nd,  (A)  and  the  1st  (B),  I  have  the 
following 

(A)  OPERATION.  (B) 

2x^  —12^2  +  21x— 10)  J7^  —    8x^  +  21^72—  20^  +  4 
2 

fCJ 2x*~  Wx^  +  42^2  —  40^-  -f-  8(07  —  2 

2.774  _■  12j:3  _|_  2Lr^  —  lO.r 

—  4.x '  -\-'21x'^  —  'SOx  +  8 

—  4:x^  +  24^3  _  42^  -f  20 

(D)  Keject  —  3 3^^+12.r  — 12 

(J'V j.i_^   4^  +  4" 


HIGHEST   C.  D.  95 


X'  — 

(^)                               (A) 
■  4:x  +  4)2.r3  —  12^2  +  21a:  —  10(2a;  —  4 
2^3  _    8.^-^  H-    Sx 

—  4a;^  +  rSx  —  10 

—  4.^2  +  IQx  —  IG 

(F) 

-  -  -  Keject  _  3  -  -    —    3j;  +    G             f ^j 

(GJ 

^  _  2)a;2  —  4a;  +  4(0;  - 

X-'  —  207 

—  2a;  +  4 

—  2a;  +  4 

-2 

REASONING. 

The  two  given  poljoiomials  being  an-anged  with  reference  to  x,  and 
no  common,  or  other  factor,  appearing,  I  proceed  at  once  to  determine 
by  successive  divisions  their  H.  C.  D. 

The  H,  C.  D.  cannot  be  higher  than  (A),  the  lower  of  the  two  ;  and, 
as  it  is  its  own  only  divisor  of  the  3rd  degree,  if  it  divides  (B),  it  is 
the  H.  CD.  As  2x^  is  contained  in  x*,  ^.r  times,  to  avoid  fractions 
I  multiply  (B)  by  2,  and,  since  2  is  not  a  factor  of  (A),  the  H.  C.  D. 
of  (A)  and  (B),  is  the  H.  C.  D.  of  (A)  and  (C).  Now  if  (A)  divides  (C) 
it  is  the  H.  C.  D.  Trj-ing  it,  I  find  a  remainder,  (D).  But  the  H.  C. 
D.  of  (A)  and  (C)  is  also  a  divisor  of  (D),  for  (D)  is  the  difference  be- 
tween (C)  and  (x  —  2)  times  (A),  both  of  which  are  divisible  by  the  H. 
C.  D.  of  (A)  and  (C).  The  question  is  now  reduced  to  finding  the  H. 
C.  D.  of  (A)  and  (D).  Upon  which  I  reason  exactly  as  before  upon 
(A)  and  (B).  Thus,  since  —  3  is  a  factor  of  (D),  and  not  of  (A),  it 
can  be  rejected  ;  and  the  H.  C.  D.  of  (A)  and  (D)  is  the  H.  C.  D.  of  (A) 
and  (E).  This  cannot  be  higher  than  (E) ;  and  as  (E)  is  its  own  only 
divisor  of  the  2nd  degree,  if  it  divides  (A),  it  is  the  H.  C.  D.  Trying 
it,  I  find  a  remainder,  (F).  Upon  this  remainder,  and  (E),  I  reason 
as  before,  upon  (D)  and  (A).  Thus,  the  H.  C.  D.  of  (E)  and  (A)  is 
a  divisor  of  (E)  and  (F),  since  (F)  is  the  difference  between  (A)  and  a 
multiple  (2x  —  4  times)  of  (E).  The  question  is  then  reduced  to  find- 
ing the  H.  C.  D.  of  (E)  and  (F).  •  Upon  these  I  reason  as  before,  re- 
jecting —  3,  which  is  not  a  common  factor,  and  hence,  forms  no  part 
of  the  H.  C.  D.  of  (E)  and  (F),  and  finding  by  trial  that  (G)  is  a  di- 
visor of  (E).     Therefore,  x  —  2  is  the  H.  C.  D.  of  (A)  and  (B). 

ScH. — The  pupil  should  be  careful  to  notice  that  at  each  step  in  this 
process,  we  show  that  the  H.  C.  D.  sought,  cannot  be  higher  than 
the  divisor  used.  Hence  the  divisor  which  terminates  the  vwrk,  is  the 
H.  C.  D. 


96  FACTORING. 

GENERAL  DEMONSTRATION  OF   THE    RULE    FOR    FIND- 
ING THE  H.  C.  D. 

Let  A  and  B  represent  any  two  polynomials  whose  H.  C.  D.  is 
sought. 

1st.  Arranging  A  and  B  with  reference  to  the  same  letter,  for  con- 
venience in  dividing,  and  also  to  render  common  factors  more  readily 
discernible,  if  any  common  factors  appear,  they  can  be  removed  and 
reserved  as  factors  of  the  H.  C.  D.,  since  the  H.  C.  D.  consists  of  all 
the  common  factors  of  A  and  B . 

2nd.  Having  removed  these  common  factors,  call  the  remaining  fac- 
tors C  and  D.  "We  are  now  to  ascertain  what  common  factors  there  are 
in  C  and  D,  or  to  find  their  H.  C.  D.  As  this  H.  C.  D.  consists  of  only 
the  common  factors,  we  can  reject  from  each  of  the  polynomials,  C  and 
D,  any  factors  which  are  not  common.  Having  done  this,  call  the  re- 
maining factors  E  and  F. 

3rd.  SupiDOse  polynomial  E  to  be  of  lower  degree  with  respect  to  the 
letter  of  arrangement  than  F,  (If  E  and  F  are  of  the  same  degree,  it 
is  immaterial  which  is  made  the  divisor  in  the  subsequent  process. ) 
Now,  as  E  is  its  own  only  divisor  of  its  own  degree  (Lem.  2),  if  it  di- 
vides F,  it  is  the  H.  C.  D.  of  the  two.  If,  in  attempting  to  divide  F 
by  E  to  ascertain  whether  it  is  a  divisor,  fractions  arise,  F  can  be  mul- 
tiplied by  any  number  not  a  factor  in  E  (and  E  has  no  monomial  fac- 
tor), since  the  common  factors  of  E  and  F  would  not  be  aftected  by 
the  operation.  Call  such  a  multiple  of  F,  if  necessary,  F'.  Then  the 
H.  C.  D.  of  E  and  F',  is  the  H.  C.  D.  of  E  and  F.  If,  now,  E  divides 
F',  it  is  the  H.  C.  D.  of  E  and  F.  Trying  it,  suppose  it  goes  Q  times, 
with  a  remainder,  E. 

4th.  Any  divisor  of  E  and  F'  is  a  divisor  of  R,  since  F'  —  QE  =  E, 
and  any  divisor  of  a  number  divides  any  multiple  of  that  number  (Lem. 
3),  and  a  divisor  of  two  numbers  divides  their  difference.  The  H.  C. 
D.  divides  E,  hence  it  divides  QE,  and,  as  it  also  divides  F',  it  divides 
the  difference  between  F'  and  QE,  or  E.  Therefore,  the  H,  C.  D.  of  E 
and  F',  is  also  the  H.  C.  D.  of  E  and  E. 

5th.  We  now  repeat  the  reasoning  of  the  3rd  and  dth  paragraphs 
concerning  E  and  F,  with  reference  to  E  and  E.  Thus,  E  is  by  hypo- 
thesis of  lower  degree  than  E  ;  hence,  dividing  E  by  it,  rejecting  any 
factor  not  common  to  both,  or  introducing  any  one  into  E,  which  may 
be  necessary  to  avoid  fractions,  we  ascertain  whether  E  is  a  divisor  of 
E. 

Gth.  Proceeding  thus,  till  tvv'o  numbers  are  found,  one  of  which  di- 


I 


HIGHEST   C.    D.  97 

vides  the  other,  the  last  divisor  is  the  H.  C.  D.  of  E  and  F,  since  at 
every  step  we  show  that  the  H.  C.  D.  is  a  divisor  of  the  two  numbers 
compared,  and  the  last  divisor  is  its  own  H.  D. 

7th.  Finally,  we  have  thus  found  all  the  common  factors  of  A  and 
B,  the  product  of  which  is  their  H.   C.  D.     q.  e.  d. 

EXAMPLES. 

1.  Find  the  H.  C.  D.  of  14aa;  —  8a  +  a.x^  —  lax'^  and 
IQa'^x'^  +  Qta-x"^  —  ISa^-x^,  and  give  the  reasoning  as 
above.  The  H.  G.  D.  is  ax  —  4a. 

2.  Find  the  H.  C.  D.  of  a^  +  Sa'b  +  dab^  +  h^  and  5a' 
+  5b%  giving  the  demonstration. 

The  H.  a  D.  is  a -{- b. 

3.  Find  the  H.  C.  D.  of  36a'  +  da^  —  27a^  —  18a'  and 
21a^b-  —  da%-  —  18a%^,  giving  the  demonstration. 

The  H.  a  D.  is  da*  —  9a\ 

4.  Find  the  H.  C.  D.  of  4:xy^  —  2y^  +  6x-^y  and  4:X^y 
+  8^^  —  Axy"^,  and  give  the  demonstration. 

The  K  a  D.  is   2x  +  2y. 

5.  Find  the  H.  C.  D.  of  Zx'  —  IQx^  +  15^  +  8  and  x^ 

—  2x*  —  6^3  4.  4^2  _j_  13^  ^  6. 

The  E.  a  D.  is   x^  -f  3a;2  +  3^7  -f  1. 

6.  Find   the   H.    C.    D.    of    2a^x^  —   2a'bx^y  +   2ab^x^y^ 

—  2b'^xy^  and  4:a^b^x-^y'^  —  2ab'^x^y^  —  2b'*xy\ 

The  H.  a  D.  is   2ax-^  —  2bxy. 


138.  JProh,  To  find  the  H.  0.  D.  of  three  or  more  poly- 
nomials. 

RULE. — Find  the  H.  C.  D.  of  any  two  of  the  given 
polynomials  by  one  of  the  foregoing  methods,  and  then  find 
the  h.  c.  d.  of  this  h.  c.  d.  and  one  of  the  eemaining 
polynomials,.  and  then  again  compare  this  last  h.  c.  d. 
with  another  of  the  polynomials,  and  find  their  h.  c  t). 
Continue  this  process  till  all  the  polynomials  have  been 

USED. 


yb  FACTOllING. 

Dem. — For  brevity,  call  the  several  polynomials  A,  B,  C,  D,  etc. 
Let  the  H.  C.  D.  of  A  and  B  be  represented  by  P,  whence  P  contains 
all  the  factors  common  to  A  and  B.  Finding  the  H.  C.  D.  of  P  and  C, 
let  it  be  called  P'.  P',  therefore,  contains  all  the  common  factors  of 
P  and  C  ;  and  as  P  contains  all  that  are  common  to  A  and  B,  P'  con- 
tains all  that  are  common  to  A,  B  and  C.  In  like  manner  if  P"  is  the 
H.  C.  D.  of  P'  and  D,  it  contains  all  the  common  factors  of  A,  B,  C, 

id  D,  etc.,  etc.    Q.  e.  d. 

EXAMPLES. 

1.  Find  the  H.  C.  D.  of  2x*  +  Qx^  +  Ax%  3x^  +  9^^  _j_  9^ 
+  6,  and  3^3  _|.  Sx^^  ^  5x  -\- 2. 

The  K   a  D.  is  x  -\- 2. 

2.  What  is   the   H.  C.  D.   of   lOa^  +  lOa^b"-  +  20a^b,  2a^ 

+  263,  and  4&«  +  12a''b^  +  4:a^b  +  12a63  ? 

Ans.,  2(a  +  b). 
^^^^ 

SECTION  III. 
Lowest  or  Least  Oommon  Multiple. 

139 »  Def. — In  speaking  of  decimal  numbers,  the  term  Least  Com- 
mon Multiple  is  correct,  but  not  in  speaking  of  literal  numbers.  For 
example,  the  numbers  (a-f-&)^  and  (a"^  —  6^)  are  both  contained  in 
(a-j-6)2  X  (a  —  b),  and  in  any  multiple  of  this  product,  as  m{a-{-b)^ 
(a  —  6).  But  whether  m{a  -f-  by~(a  —  6)  is  greater  or  less  than 
{a-\-b)'{a  —  b)  depends  upon  whether  a  is  greater  or  less  than  6,  and 
also  whether  m  is  greater  or  less  than  unity.  In  speaking  of  literal 
numbers,  we  should  say  Loicest  Common  Multiple,  meaning  the  multi- 
ple of  lowest  degree  with  respect  to  some  specified  letter. 


140.  JProh.  To  find  the  L.  G.  M.  of  two  or  more  numbers. 

RULE. — Take  the  LrrEKAii  number  of    the  highest   de- 
gree, OR  the  largest  decimal  number,  and  multiply  it  by 

ALL   the   factors     FOUND     IN    THE   NEXT   LOWER   WHICH   ARE    NOT 

EST  IT.     Again,  isiultiply   this  product   by    all  the  factors 

FOUND  in  the  next  LOWER  NUMBER  AND  NOT  IN  IT,  AND  SO 
CONTINUE  TILL  ALL  THE  NUMBERS  ARE  USED.  ThE  PRODUCT 
THUS   OBTAINED   IS   THE  L.   C.  M. 


LOWEST   C.    M.  99 

Dem." — Let  A,  B,  C,  D,  etc.,  represent  any  numbers  arranged  in  the 
order  of  their  degrees,  or  values.  Now,  as  A  is  its  own  L.  M.,  the 
L.  C.  M.  of  all  the  numbers  must  contain  it  as  a  factor.  But,  in  order 
to  contain  B,  the  L.  C.  M.  must  contain  all  the  factors  of  B.  Hence, 
if  there  are  any  factors  in  B  which  are  not  found  in  A,  these  must  be 
introduced.  So,  also,  if  C  contains  factors  not  found  in  A  and  B,  they 
must  be  introduced,  in  order  that  the  product  may  contain  C,  etc.,  etc. 
Now  it  is  evident  that  the  product  so  obtained,  is  the  L.  C.  M.  of  the 
several  numbers,  since  it  contains  all  the  factors  of  any  one  of  them, 
and  hence  can  be  divided  by  any  one  of  them,  and  if  any  factor  were 
removed  it  would  cease  to  be  a  multiple  of  some  one  or  more  of  the 
numbers,     q.  e.  d. 

1.  Find  the  L.  C.  M.  of  {x^  —  l),{x^-  —  1),  and  {x  +  1). 

MoDEii  Solution. — The  L.  C.  M.  must  contain  x^  —  1,  and  as  it  is 
its  own  L.  M.,  if  it  contains  all  the  factors  of  the  other  two,  it  is  the 
required  L.  C.  M.  The  factors  of  x^  —  1  are  (x  —  l)(x*  +,  ^  +  !)• 
But  this  product  does  not  contain  the  factors  of  (x^  —  1),  which  are 
[x  -f-  l)(a5  —  1).  Hence  we  must  introduce  the  factor  [x  +  1),  giving 
{x^  —  l)(x  -f- 1).  as  the  L.  C.  M.  of  x*  —  1  and  x*  —  1.  Now  as  this 
product  contains  the  third  quantity  it  is  the  L.  C.  M.  of  the  three. 

2.  Find  the  L.  C.  M.  of  {a  +  by,  a^-  —  h\  {a  —  by,  and  a^ 
+  Sa'b  +  3b-'a  +  b\ 

Suggestions. — The  last  is  (a  -\-  h)'^  which  contains  the  factors  of  the 
1st,  but  neither  of  the  factors  of  the  3rd.  Both  factors  of  (a  —  6)* 
must,  therefore,  be  introduced,  giving  [a  -\-b)'^{a  —  b}''^  as  the  L.  G. 
M.  of  the  1st,  3rd  and  4th.  And  as  it  contains  both  factors  of 
a^  —h^,  viz. ;  (a  +  b)(a  ~  6),  it  is  the  L.  C.  M.  of  all. 

3.  Find   the  L.   C.    M.    of    {x^-  —   4a2),    {x  +  2a)^  and 
{x  —  2a)3.  L.  a  M.,  {x^  —  4a-^)3. 

4  Find  the  L.  C.  M.  of  x-^  +  2xy  +  y^  and  x^  —  xyK 

L.  a  if.,  {x  +  y){x^  —  xy^). 

5.  Find  the  L.  C.  M.  of  l^a-ma;^,  ^^a^bm'^,  and  Ib^x"^. 

L.  a  M.,  10a^b^m-x\ 

6.  Find  the  L.  C.  M.  of  lGx^y\  10a"-x,  and  ^Oa^x^. 

L.  G.  31.,  80a^x^y\ 

ScH. — In  applying  this  rule,  if  the  common  factors  of  the  two  num- 
bers are  not  readily  discerned,  apply  the  method.of  finding  the  H.  C.  D., 
in  order  to  discover  them. 


100  ^ACTOEING. 

7.  Find  the  L.  C.  M.  of  x^  —  2ax'i  +  4:a-x  —  %a\  x^-{-  '2ax^ 
+  4a2^  +  8a^,  and  x"^ —  4a-. 

Model  Solution. — The  L.  C.  M.  of  tliese  numbers  must  contain 
x^  —  2ax^  -\-  4:a^x  —  8a'^ ;  and  as  it  is  its  own  L.  M.,  if  it  contains  all 
the  factors  of  x^  -\-  2ax^  -}-  ia^x  -\-  8a'^,  it  is  the  L.  C.  M.  of  these  two 
polynomials.  But  as  the  common  factors  of  these  numbers,  if  they 
have  any,  are  not  readily  discerned,  we  apply  the  method  of  H.  C.  D. 
and  find  that  x^  -)-  4a^  is  the  H.  C.  D.  of  the  two.  Since,  then,  a;* 
—  2ax^  -j-  4ia^x  —  8a^  contains  the  factor  x'  -j-  4m^  of  the  second 
number,  it  is  only  necessary  to  introduce  the  other  factor  in  order  to 
have  the  L.  C.  M.  of  the  two.  Now  (x^  +  2ax2  +  4:a^x  -f  8a^)  4- 
{x^  +  4a«)  =x-\-2a.  Hence  (x^  —  2ax^  +  4m^x  —  8a^){x  +  2a)  or 
X*  —  16a'*  is  the  L.  C.  M.  of  the  first  two  numbers,  since  it  contains  all 
the  factors  of  each,  and  no  more.  Now,  to  find  whether  x'*  —  16a*  is 
a  multiple  of  the  remaining  number,  x^  —  4a^ ,  or,  if  it  is  not,  what 
factors  must  be  introduced  to  make  it  so,  we  proceed  in  the  same  way 
as  with  the  first  two  numbers.  But  our  first  step  (or  124)  shows  us 
that  X*  —  16a*  is  a  multiple  of  x*  —  4a*.  .  • .  x*  —  16a''  is  the  L.  C. 
M.  of  the  three  given  numbers* 

8.  Find  the  L.  C.  M.  oi  Gx"- -~  x  —  1  and  2x^  +  Sot  —  2. 

L.  a  M.,  (2a;2  4-  3^  —  2)  (3^  +  1). 

9.  Find  the  L.   C.  M.   of  x^  —  ^x^  +  23a;  —  15  and 

x-^  —  8^7  +  7. 
L.  a  M.,  {x^  —  dx^  +  2307  —  15)(«;  _  7)  =  o:^  —  16^:3 
+  86x^  —  llGx  +  105. 

10.  Find  the  L.  C.  IVL  of  2x  —  1,  4:X^  —  1,  and  Ax^  -f  1. 

11.  Find  the  L.  C.  M.  of  x^  —  x,  x^  —  1,  and  x^  +  1. 

L.  a  31.,  x{x<^  —  l)=x-'  —  x. 

12.  Find  the  L.  C.  M.  of  x^  —  6x^  +  llo;  —  6,  x^  —  9x^ 
+  2607  —  24,  and  x^  —  8x^  +  Idx  —  12. 

L.  a  J/.,  {x  —  l){x  —  2){x  —  3){x  —  4.)  = 
X*  —  10o;3  +  35o7»  —  5007  +  24. 


I 


SYNOPSIS. 


101 


Synopsis  for  Eeview, 


f  Common. 


Factor.  — Div  isor.  — Meas  ure. 


a 
[h. 


Q*  p'  j-  Distinction, 


i  Common. 
MuUiple.    \^^10.M^^    [Distinction. 

i  Composite. 
Numbers.  <  Prime. 

(  Prime  to  each  other. 

^  Scholium. — How  these  terms  are  applied. 


Prop.  1.  A  monomial.     Dem. 

Prop.  2.  A  poly,  with  mon.  factors. 

Prop.  3.  «2  +2ab-\-h'\     Dem. 

Prop.  4.  a^  —  h^.     Dem. 

Prop.  5.  One  factor  given.     Dem. 

Prop.  6.  (a"*  ±  h-^)^{a  ±b).    Dem.  - 


Dem. 


Sch. 


f^ 


ft 


Prop.  7.  A  trinomial.     When  ? 
Prop.  8.   Separating  into  parts. 


Form  of  quo- 
tient. 
Cor.  Frac.        and 
[  neg.  exp'ts. 

How  ?    Dem. 


Distinction  between  G.  C.  D.  and  H.  C.  D. 

ri.  Dem. 

Lemmas.      \  \  ^JJ)^; 
[  4.  De')n. 
General  Method. — Prob.  1.    Rux-e.     Dem. 
Prob.  2.  Of  more  than  2  numbers.    RuiiE.     Dem. 


^  i  Distinction  between  Least  and  Lowest  C.  M. 

Q  ■<  Prob.     Rule.     Dem. 

^   (  Scholium.  By  means  of  H.  C.  D. 


Test  Questions. — What  are  the  factors  of  a*  —  b^  1  Of  a*  -}"  2a& 
+  62  ?  Of  a*  —  2ah  +  ?>2  ?  Of  1  —  2ic  +  .x^  ?  Of  Sa"  +  Ga^x 
+  Sa^x"?  Of  x*  +  1/2+2x2/?  Of  x^  —  x  —  12?  Of  m?  Of 
a^h^yl  Of  x3  —  y^  ?  State  the  general  rule  for  testing  the  divisibil- 
ity of  the  sum  or  difference  of  like  powers.  Prove  one  of  the  cases, 
as  (a™  —  5'^' )  +  (a  —  h).  Distinction  between  H.  C.  D.  and  G.  C.  D. 
Between  Lowest  and  Least  C.  M.  Explain  the  process  of  finding  each 
by  factoring.  Give  the  General  Eule  in  each  case,  and  its  demon- 
stration. 


102  FKACTIONS. 

OHAPTEE    III. 

FRACTIONS. 

SUCTION  I. 
Definitions  and  Fundamental  Principles. 

141,  A.  FractioUf  in  the  literal  notation,  is  to  be 
considered  as  an  indicated  operation  in  Division.  It  is 
written,  as  in  common  arithmetic,  with  one  number  above 
another  and  a  line  between  them.  The  number  above  the 
line,  i.  e.,  the  dividend,  is  called  the  Numerator;  and  the 
number  below  the  Hne,  i.  e.,  the  divisor,  is  called  the  Denom- 
inator. 

^,         2a  —  5m.x^  Numerator  or  Dividend. 
Thus: p- 

3c  +  4oX'^     Denominator  or  Divisor. 
This  expression  means  nothing  more  than  (2a  —  5mx^) 

-^  (3c  +  4^^). 

Taken  together  numerator  and  denominator  are  called 
the  Terms  of  the  Fraction. 

142.  ScH. — In  the  literal  notation  it  becomes  impracticable  to 
consider  the  denominator  as  indicating  the  number  of  equal  parts  into 
which  unity  is  divided,  and  the  numerator  as  indicating  the  number 
of  those  parts  represented  by  the  fraction,  since  the  very  genius  of  this 
notation  requires  that  the  letters  be  not  restricted  in  their  signification. 

Thus  in  -,  it  will  not  do  to  say,   h  represents  the  number  of  equal 

parts  into  which  unity  is  divided,  since  the  notation  requires  that 
whatever  conception  we  take  of  these  qiiantities  should  be  sufficiently 
comprehensive  to  include  all  values.  Hence  b  may  be  a  mixed  num- 
ber. Now  suppose  b  =  4|.  It  is  absurd  to  speak  of  unity  as  divided 
into  4|  equal  parts. 


DEFINITIONS.  103 

143,  The  Value  of  a  Fraction  is  the  quotient  of 
the  numerator  divided  by  the  denominator. 

14:4:,  Cor.  1. — Since  nuraerator  is  dividend  and  denom-' 
inator  divisor,  it  follows  from  {100 ,  101^  102)  that  divid- 
ing or  muUijjlying  both  terms  of  a  fraction  by  the  same  quan- 
tity does  not  alter  its  value ;  that  multiplying  or  dividing  the 
numerator  multijMes  or  divides  the  value  of  the  fraction  ;  and 
that  multiplying  or  dividing  the  denominator  divides  or  multi- 
plies the  fraction. 

146,  CoE.  2. — A  fraction  is  multiplied  by  its  denominator 

by  simply  removing  it.     Thus,  to  multiply  j  by  4  it  is  only 

necessary  to  remove  the  denominator,  which  gives  3.     This 
is  the  same  as  dividing  the  denominator  by  4,  for  that  gives 

-,  or  3,  as  the  product.    So  also  to  multiply  -  by   ^    is    to 

drop  the  denominator,  x,  which  gives  a.     It  is  evident  that 
a  is  ^  times  as  much  without  being  divided  by  x  as  when 

a 
divided  :  that  is,  a  is  ^  times  -. 

X 

146,  The  terms  Integer  or  Entire  Number,  Mixed  Num- 
ber, Proper  and  Improper,  are  applied  to  literal  numbers, 
but  not  with  strict  propriety.  Thus,  whether  m  +  n  is  an 
integer,  a  mixed  number,  or  a  fraction,  depends  upon  the 
values  of  m  and  n,  which  the  genius  of  the  literal  notation 
requires  to  be  understood  as  perfectly  general,  until  some 
restriction  is  imposed. 

For  convenience,  we  adopt  the  following  definitions  : 

147,  A  number  not  having  the  fractional /orm  is  said 
to  have  the  Integral  Form  ;  as  m  -\-  n,  2c^d  —  8a~'^x 
+  ^x^y\ 

14S,  A  polynomial  having  part  of  its  terms  in  the  frac- 
tional and  part  in  the  integral  form,  is  called  a  3Iixed 

JS^uniber;  as  a  —  .r  +  ^^^  """  ^•^_. 

am- 


104  FEACTIONS. 

14:9,  A  Proper  Fraction,  in  the  literal  notation, 
is  an  expression  wholly  in  the  fractional  form,  and  which 
cannot  be  expressed  in  the  integral  form  without  negative 
exponents. 

By  calling  such  an  expression  a  proper  fraction,  we  do 
not  assert  anything  with  reference  to  its  value  as  compared 

with  unity.     Thus  -  is  a  proper  fraction,  though  it  may 

be  greater  or  less  than  unity.     It  may  also  be  written  ab  ~\ 

150,  An  Improper  Fraction  is  an  expression  in 
the  fractional  form,  but  which  can  be  expressed  in  the 
integral  or  mixed  form  without  the  use  of  negative  expo- 

nents.     Thus,  --, —  =  2a -— r  :    the  former 

n  —  6ab  n  —  Sao 

of  which  is  called  an  improper  fraction. 

151,  A  Simple  Fraction  is  a  single  fraction  with 

2^ ■^^ 

both  terms   in  the   integral  form.     Thus  — - — -— r— ,,  and 
^  4y  +  2cc? 

2cm^2  •      1    i?      i.- 

• — -  are  simple  fractions. 

x  —  'ly 

152,  A  Comxyouncl  Fraction  is  two  or  more 
fractions  connected  by  the  word  of;  but  the  expression  is 
not  generally  applicable  in  the  literal  notation.     Thus  we 

may  write  t^  of  -r  with  propriety,  but  not  -  of  -,    unless  a 
34  xj.*'  on 

and  h  are  integral,  so  that  the  fraction  -  may  be  considered 

2 
as  representing  equal  parts  of  unity,   as  -  does.      If    the 

o 

word  of  is  considered  as  simply  an  equivalent  for  X,  the 
notation  is,  of  course,  always  admissible.  But  it  is  scarcely 
a  simple  equivalent. 


FUNDAMENTAL  PRINCIPLES.  105 

153.  A  Complex  Fraction  is  a  fraction  having  in 
one  or  both  its  terms  an  expression  of  the  fractional  form. 

c  ma^ 

^m  —  -  —J- 

Thus ,  and are  complex  fractions. 

5  -  —  a 

y 

lo4:,  A  fraction  is  in  its  Lowest  Terms  when  there  is  no 
common  integral  factor  in  both  its  terms. 

lSo»  The  Loivest  Common  Denominator  is 

the  number  of  lowest  degree,  which  can  form  the  denomi- 
nator of  several  given  fractions,  giving  equivalent  fractions 
of  the  same  values  respectively,  while  the  numerators  re- 
tain the  integral  form. 

156,  Hedlictionf  in   mathematics,  is   changing   the 
form  of  an  expression  without  changing  its  value. 


SIGNS  OF  A  FRACTION. 

lo7»  In  considering  the  signs  of  a  fraction,  we  have  to 
notice  three  things,  viz.;  the  sign  of  the  numerator,  the  sign 
of  the  denominator,  and  the  sign  before  the  fraction  as  a 
whole.  This  latter  sign  does  not  belong  to  either  the  nu- 
merator or  denominator  separately,  but  to  the  whole  ex- 
pression. Thus,  in  the  expression  —  — H^ — _,  in  the  nu- 
^  2x-{-  4?/^ 

merator  the  sign  of  4a  is  -{-,  and  of  5cd  — .  In  the  denom- 
inator, the  sign  of  2x  is  -\-,  and  of  4:y-  -\-  also.  The 
sign  of  the  fraction  is,  — .  These  are  the  signs  of  ope- 
ration.    {SOf  33 f  34,) 

158,  The  essential  character  of  a  fraction,  as 
positive  or  negative,  can  only  be  determined  when  the  essen- 
tial character  of  all  the  numbers  entering  into  it  is 
known.  It  may  then  be  determined  by  principles  already 
given.     {86,  106,) 


106  FEAGTIONS. 


EXAMPLES. 


1.  Is  the  fraction — —    essentially    positive,    or 

negative,  when  a,  m,  x,  and  y  are  each  negative  ? 

Model  Solution. — Since  ( —  a)^  =:  a^,  4a^  is  essentially  positive. 
Since  ( —  m)( —  x)=  mx,  the  term  3mx,  in  itself,  is  positive,  and  the  nu- 
merator becomes  4a ^  —  (-f  Smx),  or  4,a^  —  Smx  {7S).  Now,  whether 
4a^  —  Smx  gives  a  -f-  or  a  —  result,  depends  upon  the  numerical  val- 
ues of  a,  m,  and  x.  K  4a^  ^  Swicc,  4a^  — Smx  is  -\-  ;  but,  if  4a'^<^ 
Smx,  4a^  —  3?nx  is  — .  Again,  since  ( —  x)'^  =  —  ic•^  the  first  term  of 
the  denominator,  2ic^,  is  essentially  negative.  And  since  ( —  yY  =2/^' 
the  term  4?/^  is  essentially  positive  and  the  denominator  becomes 
—  2x"*  +  (4-  4^/2),  or  —  2x^  +  4ty^.  Whether  this  is  +  or  — ,  de- 
pends upon  the  relative  values  of  x  and  y.  K  we  suppose  4a2  >>  Smx 
the  numerator  becomes  -\-,  and  if  2x^  be  greater  than  4^/^  the  denom- 
inator becomes  — ,  and  we  have ,  which  gives  a  positive  result. 

2a'^x-  -f-  4:1/^ 

2.  What  is  the  essential  sie^n  of  —--, ,  '  ,  when    a  = 

^  3b"-  —  Ax 

—  S,  X  =  —  2,  2/  =  —  4,  and  b  =  —  5?     Ans.,  — . 

.  ^x^y  -}-  CLX'^ 

3.  What    is  the  essential  sign  of ~ ,    when 

oCL^X  „ 

X  =  —  2,  fl  =  —  1,  and  ?/  =  3  ?  Ans.,  — . 

4.  What  is  the  essential  siefn  of ^ ,  when 

^  —  2y'  —  3cx 

X  =^  4:,  a  =  —  2,  y=  —  1,  and  c  =  Q^       Ans.,  -f . 


SECTION  IL 
Eeductions. 


159.  There  are  five  principal  reductions  required  in 
operating  with  fractions,  viz. ;  To  Lowest  Terms, — From 
Improjjer  Fractions  to  Integral  or  Mixed  forms, — From  In- 
tegral or  Mixed  Forms  to  Improper  Fractions, — To  forms 
having  a  Common  Denominator, — and  from  the  Comi)lex  to 
the  Simple  Form, 


BEDUCTIONS.  107 

I^VOh.  !•   To  reduce  a  fraction  to  its  loioest  terms. 

R  TILE. — Reject  all   common  factors  from   both  terms  ; 

OR    DFSmDE   BOTH    TERMS   BY    THEIR    H.   C.  D. 

Dem. — Since  the  numerator  is  tlie  dividend  and  the  denominator  the 
divisor,  rejecting  the  same  factors  from  each  does  not  alter  the  value 
of  the  fraction.  {100.)     Having  rejected  all  the  common  factors,   oi*,  , 
what  is  the  same  thing,  the  H.  C.  D.  (which  contains  all  the  common 
factors),  the  fraction  is  in  its  lowest  terms.  (154:.) 

EXAMPLES. 

1.  Reduce — -— to  its  lowest  terms. 

Model  Solution. — Resolving  the  terms  of  the  fraction  into  their 

prune  factors,  I  have  ; ; ^ =  — <^ — -. 

'da*  -f-  ija-^x  -\-  '6a^x^         ^JX-  a(a..^>f=^)[a  -\-  x) 

Now,  cancelling  the  common  factors,  3,  a,  and  a  -\-  x,  which  is  divid- 
ing dividend  and  divisor  by  the  same  quantity  and  hence  does  not 

alter  the  value  of  the  fi-action,  I  have  — ,  or  — ■ .     Since 

a{a  -f-  X)  a'^  -j-   ax 

in  this  there  is  no  factor  common  to  numerator  and  denominator,  it  is 

in  its  lowest  terms.  {154:,) 

2.  Reduce  — -,  -— ^,  — ,  and  — to  their 

Ibcx'^y^  55i)m^xy=    dba^x^  15a^c^x 

lowest  terms. 

Results  (not  given  in  order),  - —  ,  — ,  Za-x  and  — . 

Qox'^     omy~  bcx^y 

o    r»   n         a  —  X     ax  -{-  x^  .  2x^  —  16^  —  6  ,    ^.    . 

3.  Reduce ,  — — ,  and — ,  to  their 

a2  —  x-^  ab-^  4-  t^x  3x^  —  24^  —  9 


lowest  terms. 


X    2 
Results  (not  in  order),  — ,  -,  and 


6= '  3'  a-{-x 

.    T>   ,        n2  —  2n  +  1   3a3  —  Sab^        ^      x'^  —  ¥x 

4.  Reduce , ■  and to 

71^  —  1      '   5ab  +  56^ '  x-^  +  2bx  +  h^ 

their  lowest  terms. 

T,      7,    ^-  —  bx  n  —  1        ,3^2  —  Sab 
Results, — -,  — — -,  and . 

X  -{-  b     n  -]-  1  56 


108  FRACTIONS 

—  2xy-\ 
xi  —  2/" 


/>»2 ^orxi  -\-  11'^ 

5.  Reduce —  to  its  lowest  terms. 


X 7/ 

Result, ~. 

X  -^  y 

6.  Reduce '- — — r —  to  its  lowest  terms. 

3x^  +  t.y+Sr  BesulL^f^. 


X  -i-  y 

^    ^   -        5w2  —  lOn  +  5  ^    .^    , 

7.  Reduce to  its  lowest  terms.  ^, 

7n^'  — 7  5(n  — 1) 

^''^^'^  tT^TTT)- 

8.  Reduce ^  to  its  lowest  terms. 

y2  —  2xu  4-  X  ,  , 

Result,  ^  ^-  ■^^ 

y  —  X 
72,3  —  2n2 

9.  Reduce r  to  its  lowest  terms. 

n2  —  4/1  +  4  -r,      7         ^' 

Result^  -. 

n  —  2 

10.  Reduce '-  to  its  lowest  terms. 

^'  —  ^^                                          o^  +  tf^a:2  4-  X* 
Result.  -~ . 

a"  -H  •^■- 

ScH.  1. — Since  the  H.  C,  D.  is  the  product  of  all  the  common  fac- 
tors {116),  the  above  process  is  equivalent  to  dividing  both  terms  of 
the  fraction  by  their  H.  C.  D.  Whenever  the  common  factors  of  the 
terms  are  not  readily  discernible,  the  process  for  finding  their  H.  C. 
B.  {137),  niay  be  resorted  to. 

-.  ^     T^    1         ^^  —  «^  «^  —  h^  x^  -\-  y^        _    fls  —  53 

11.  Reduce , r-, —,  and  r   ^^  their 

x^  —  a*   a  —  b  x^  —  y^  a*  —  a-b'^ 

lowest  terms. 

^      ,  ,      ,     x^  —  xy  4-  y^  a^  -{-  ab  -\-  b^        ,       1 

Results,  a2  +  a6  +  b\ '  , — ,  and  ■ . 

'        X  —  y  a^  +  a^b  x*^a'' 

,^    ^    ,         6^2  _   7^  _   20   3a^2  _  lOax  +   3a 

12.  Reduce  — — — — , -7—.   and 

4a;3  —   27^;   +   5    ba'x-^  —  ba'^x  —  BOa^* 

24^^  —  22^2  +  5 

to  their  lowest  terms. 


48a;i  -4-  l^x^  —  15 


„      ,^     2j;2  —  1      3^  —  1  ,        3^  +  4 

Results, -,   ■— — ,  and 


4.07=^  +  3    10a  +  5ax  2x'^  +  5x  —  1 


EEDUCTIONS.  109 

ScH.  2. — The  opposite  process  is  sometimes  serviceable,  viz  :  the 
introduction  of  a  factor  into  both  terms  of  a  fraction  which  will  give 
it  a  more  convenient  form. 

13.  What  factor  will  chanofe to  ? 

Ans.,  X  —  1. 

14.  What  factor  will  chanefe ; to ? 

°  a  —  b  a2  —  ^2 

Ans.,  a  -\-  h. 

15.  What  factor  will  change   — o^o^   ■    a      :nT  *o 


a»  —  lijx* 


a3  — 

-  2a-x  +  4aa;2 

—  Sx^ 

Ans., 

a  +  2x, 

a'  - 

-  2a'^b  +  4«52 

—  Sb' 

16.  What   factor   will   chansfe   — ; to 

[Note. — It  requires  no  special  ingenuity  to  solve  such  problems, 
since,  if  the  factor  does  not  readily  appear,  it  can  be  found  by  divid- 
ing a  term  of  one  fraction  by  the  corresponding  term  of  the  other.  ] 


160*  I*rob,  2.  To  reduce  a  fraction  from  an  im- 
proper  to  an  integral  or  mixed  form. 

MULE. — Perform  the  division  indicated.   (141,) 
Dem. — The  operation  is  explained  in  the  same  manner  as  the  cor- 
responding case  in  division. 

EXAMPLES. 

1.  Keduce  — '■ — r to  an  intef^ral  or  mixed  form. 

2x 

Model  Solution. — This  being  an  indicated  operation  in  Division,  I 
have  but  to  perform  the  division.  Now,  since  the  sum  of  the  quo- 
tients is  equal  to  the  quotient  of  the  sum,  I  have  but  to  divide  each 
term  of  lOax  -\-  c  —  l!>  by  2x,  or  divide  such  as  I  can  and  indicate  the 
division  of  the  others,  and  add  the  results  {112).     Thus  I  find  that 

i r=  5a  +  -^ . 

2x  2x 


110  FBACTIONS. 

^    ^    _  3a'  —  9ac  -\-  X  —  Ga    5xy   -{-   ab   -{-    .x 

2.  lieduce ,  '■ ,    and 

Sa  X 

15^2  _  4^  +  6 

to  integral  or  mixed  forms. 

oa 

4(2  —  6  ah  X 

Besults,  3a ,  5y  +  1  H ,  and  a  —  3c  —  2-\ — - . 

oa  X  "da 

^    _    ^        x''  +  12.r  +  18    a2  +4o6  +  462  +c  j?^  —  -^3 

o.  Keduce ; , ,  ^~  and 

X  -{-  6  a  -\-  zh  X  -\-  y 

to  inteoral  or  mixed  forms. 

a  -i-  X  ^ 

2^3                           9 
Besults,  x^  —  j;v  +  V^ — >  ^  +  9 or 

6  +2x ^,  and  a  +  26  +        ^ 


^  +  3  a  -\-2b 

,    ^    ^         a3  —  1     ^5  —  y5  a^  ^  3^96  _(-  :-3Qr?)2  ^  53 

4.  Keduce  — ,  --, — ; ,  and 

a  —  1       X —  y  a-^  +  2a6  +  62 

to  intef^ral  or  mixed  forms. 

x-^y 

Besult,  a  -{-  h,  X*  -{-  x^y  +  ^-y^  +  ^y^  +  2/^  x^  —  x"y 

+  xy^  —  y^,  and  a^  -f  a  +  1. 

1S1»  Cor. — By  means  of  negative  indices  {exponents)  any 
fraction  can  he  expressed  in  the  integral  form. 

EXAMPLES. 

5.  Express  — ; -; — -  in  the  inteeral  form. 

^         m{a-\-  b)~^  ^ 

Model  Solution.    ^- =   {a  4-  h)  X  -  X . 

But-  =  m  — ^,    and  -— =(a4-h)^.      Hence — 1=: 

(a  4-  h){a  +  &)'w-i  =  (a  +  5)'  m-\  ora-^m-^  -f  3a26w^-^4- 
6.  Express  , — —,  -, --,  — r-^-— — ,  and  — -^ ■ — --   in 

Sx^y     {x -\- y)^   a~^x~^y^  {m  —  ?i)-^ 

tlie  integral  form. 

Besults,  {x  +  2/)~ ',  a-x^y,  m^x^  —  m^nx'^  —  m.n^x^  -f-  n?x^ 
and  7  X  8~^a;~Y 


REDUCTIONS.  11 1 

1G2.  Prob,  3. — To  reduce  numbers  from  the  integral  or 
mixed  to  the  fractional  form. 

RULE. — MuiiTiPLY  t:he  integbal  part  by   the  given  de- 

KOailNATOK,  AND  ANNEXING  THE  NUMERATOR  OF  THE  ERAC- 
TIONAIi  PART,  IF  ANY,  WRITE  THE  SUM  OVER  THE  GIVEN 
DENOMINATOR. 

Dem. — In  the  case  of  a  number  in  the  integral  form,  the  process 
consists  of  multiplying  the  given  number  by  the  given  denominator 
and  indicating  the  division  of  the  product  by  the  same  number,  and 
hence  is  equivalent  to  multiplying  and  dividing  by  the  same  quantity, 
which  does  not  change  the  value  of  the  number.  Tile  same  is  true  as 
far  as  relates  to  the  integral  part  of  a  mixed  form,  after  which  the  two 
fractional  parts  are  to  be  added  together.  As  they  have  the  same  di- 
visors, the  dividends  can  be  added  upon  the  principle  that  the  sum  of 
the  quotients  equals  the  quotient  of  the  sum  {103). 

EXAMPLES. 

1.  Keduce  2a  —  x-  +  — to  a  fractional  form. 

a  —  X 

Model  Solution. — Multiplying  2a  —  a^  by  a  —  x,  I  have   2a*  — 

ax*  —  2aa;  -j-  x^,  which  divided  by  a  —  x,  of  course  equals  2a  —  a*  ;  or 

2a*  —  ax*  —2ax-\-x^  „     „  ,  3ax  —  4a* 

2a  —  ic*  = ! .  .-.  2a— x*  H = 

a  —  X  'a  —  X 

2a*  —  ax*  —  2ax  +  x^,  3ax  —  4a*   _  . 

h  .  But,  as  the  sum  of  these  two 

a  —  X  a  —  X 

quotients  equals  the  quotient  of  the  sum,  I  have,   after  uniting  eim- 

.,      ^             x^  —  ax*  4-  ax  —  2a^ 
liar  terms, ■ . 


62 

2.  Eeduce  a  —  h  4- to  the  form  of  a  fraction. 

a-\-b  ^      ^        a^ 

Mesult, 


a  +  b 

1x 

3.  Beduce  1  +  to  the  form  of  a  fraction. 

y  —  ^  y  4.  X 

Result,  y-JL±, 

y  —  X 

4.  Reduce  a-\-b —  to  the  form  of  a  fraction. 

«  — ^  b  b 


Result, r  or 


b       b  —  a 


112  FRACTIONS. 

5.  Reduce  ^  +  2  + —r. —  to  the  form  of  a  fraction. 

Mesuit,  -. 

X —  2 

2x 5 

6.  Reduce  ^x —  to  the  form  of  a  fraction. 

Result, 


X  A-  \ 
7.  Reduce  x  -\- 1  -\-  —^—  to  the  form  of  a  fraction. 


Result, 

la^^hc  4-  9< 
8.  Reduce  2a  —  36  +  4c  + 


3 

Q. 


X 

2a"^bc  +  9a62c  —  12ahc^ 


^«^^      Result,  fa. 


^    ^    ,                 2abx  —  2b^x  ,        ,      .. 
9.  Reduce  x ■. 7 —  to  a  fraction. 

a  +  b 

x'i v^  +  7  7 

10.  Reduce  a:  +  y  — Result, . 

^  X  —  y  y  —  X 

x^  —  ^x'iy  +  ^xy^  —  y^ 

11.  Reduce  x'^  +  2xy  +  1/2 ^   ,        ' • 

X  -\-y 

12.  Reduce  -^ {a^x^  +  oPx).  Result, - 

a'i  —  x'^  d'  —  ^ 

13.  Reduce  3a  —  9 —7—.  Result, 


a  +  3    '  '    aH-3 


163.  JProb,  4,  To  reduce  fractions  having  different 
denominators  to  equivalent  fractions  liaving  a  common  de* 
nominator. 

RULE. — Multiply  both  terms  of  each  fraction  by  the 

DENOMINATORS  OF  ALL  THE  OTHER  FRACTIONS. 

Dem. — This  gives  a  common  denominator,  because  each  denomina- 
tor is  the  product  of  all  the  denominators  of  the  several  fractions. 


REDUCTIONS.  113 

The  value  of  any  one  of  the  fractions  is  not  changed,  because  both  nu- 
merator and  denominator  are  multiplied  by  the  same  number  {100). 

EXAMPLES. 

X   2 b         3a X 

1.  Eeduce  the  fractions  -? j,  and 7  to  equivalent 

y  a  +  b  a  —  0 

fractions  having  a  common  denominator. 

Model  Solution.  — Multiplying  both  terms  of  the  fraction   —  by  a  -f  & 

and  a  —  5,  or  by  a*  —  h^,   I  have  -5 — —  which  has   the   same 

value  as  — ,  since  the  numerator  and  denominator  have  been  multiplied 
by  the  same  number.      In  like  manner  mtdtipljdng   both  terms  of 

6  ,  ,  ,     T  ,  2av  —  ahiJ  —  2by  4-  b^v   .-. 

-  by  V  and  a   —  b,   I  have  — : -^ —- -,  the  value  of 


a  A-  b  a^v  —  b'-y 

2-6 
■which  is  the  same  as         ■    ,  since,    etc.      Finally,    multiplying    both 

„  3a  —  X  ,  ,        ,    ,     ^  ,         3a^  y  —  axy  4-  ^aby  —  bxy 

terms  of ,-?  by  y  and  a  +  ^.   I  bave r^ r^ '-, 

a  —  b^  a^y  —  b^y 

which  has  the  same  value  as  —  since,  etc.     These  fractions  have 

a  —  6 
the  common  denominator  a^y  —  b'^^y  as  in  each  case  the  new  denom- 
inator is  the  product  of  all  the  old  ones. 

2.  Reduce   ■^,  ■—,  J^  and  —  to  forms  having  a  C.  D. 

2?/    2c   Ix  ?i2  *= 

3x 
Queries. — By  what  are  both  tei-ms  of  --—   to*  be    multiplied  ?     By 

what  both  terms  of  jr—  ?     By  what  both  terms  of  — ^? 

42c?i-j:2    'JOhn^xy   12cn-y^         ,  2^cmxy 
'  2Scn'Xy'  28cn^xy*  28cn^xy  28cn^xy' 

rp       rp    _1_    1  1     rj* 

3.  Reduce  -,  — - — ,  and  to  forms  having  a  C.  D. 

ij  O  J-     ~t~    X 

^      ,      5a;  +  5^2    3  4-  6jt  +8a;2        ,  15  —  Ihx 

Results,  -— -— -,     ^  ^    ,    .,  ^ ,  and.  -,  _    ,    -, .,  . 

15  +  15j7      15  +  Ibx  15  +  15j; 

4.  Reduce r,  and r  to  forms  having  a  C.  D. 

a  +  6  a  —  h 

^      ^^    a^'  —  ab        .  ab  +  h^ 

JtiesuUs, —,  and 7-- 

a*  —  b-  a-  —  0' 


114  FRACTIONS. 

ScH.  — Practically,  this  method  consists  in  multiplying  all  the  denom- 
inators together  for  a  new  denominator,  and  each  numerator  into  all 
the  denominators  except  its  own  for  a  new  numerator.  But  it  is  much 
better  to  repeat  the  rule  as  given  above,  and  let  that  be  the  form  of 
conception,  as  it  keeps  the  principle  constantly  before  the  mind. 

164,  Cor.  To  reduce  fractions  to  equivalent  ones  having 
the  Loioest  Common  Denominator ,  find  the  L.  CM.  of  all  the 
denominators  for  the  netv  denominator.  Then  multiply 
both  terms  of  each  fraction  hy  the  quotient  of  that  L.  CM. 
divided  hy  the  denominator  of  that  fraction. 

Dem.— The  purpose  in  getting  the  L.  C.  M.  is  to  get  the  lowest  num- 
ber which  can  be  divided  by  each  of  the  denominators.  That  the  pro- 
cess does  not  change  the  value  of  the  fractions  is  evident  from  {100), 
the  same  as  under  the  general  rule. 

EXAMPLES. 

a  a^  a^ 

5.  Eeduce ? r->  and  7- ^>  to  equivalent  frac 

1  —  a    (1  —  «)2  (1  —  ay         ^ 

tions  haA' ing  the  L.  C.  D. 

MODEL   SOLUTION. 

Opebation.— The  L.   C.  M.  of  1  —  a,    (1  —  a)«,  and  (1  —  a)^  ip 

3        a       X  (1  —  «)"  ^       a  —  2a^  +  a'  a'  X  (1  —  a) 

^         ^^  *    (1— a)X(i— aj^        1  — 3a  +  3a^  —  a^'(l— a)2x(l— a) 

a^  —  a^  _       a*  a* 

and 


^  ^nr 


1  —  '6a -{-'6a^  —  a-*'         (i  — a;-^       1— ;ia-}-3a    — a 

Explanation. — ^By  inspection  I  observe  that  (1  —  a)^  is  the  L.  C.  M. 
of  the  denominators,  since  it  is  the  lowest  number  which  contains  it- 
self, and  it  also  contains  each  of  the  other  denominators.     Now,  to 

make  the  denominator  of   ,  (1  —  a)^,    I    must    multiply  it    by 

(1  —  a)^  4-  (1  —  a)  ;  i.  e.,  by  (1  —  a)'.     But  to  preserve  the  value  of 
the  fraction,  I  must  multiply  the  numerator  by  the   same  quantity. 

Thus         "      .    a(l-°l'  «    -2a^+a-'     .,,._,,, 

1 — a         (1 — a)'*        1  —  3a-\-'6a'^ — a-* 

6.  Eeduce —> —to  forms  having  the  L.  0.  D. 

x~y      x  +  y 

x^—xif—  xHi  +  ?/'        ,   x^+  xHj  +  xip  -f  y' 
R,sulis,  ;^._  y/  and ^^^i -. 


REDUCTIONS.  115 

7.  Reduce >  — > >  to  forms  having  the  L.  C.  D. 

x+y  x^  -\-  y^  X  +  y 

„      -^  a       i{x^-xy+y^)        ,  c{x-'— xy-\-y^) 

ResuUsj J  - — T-r—r-^^  and  -^ -^  /  ' , 

x^+  y^         x^+  y^  x^  -h  y^ 

8.  Eeduce  nm, ? ?  to  forms  having  the  L.  C.  1). 

m  +  n    7)1  —  u 

Suggestion. — Regard  mii  as  — -. 

„      ,,     m^7i  —  mn^  (m  —  nY       ,  (7^  +  ny 
Results f ?^^- f-jand-^^— '-. 

9.  Eeduce > 1?  - — ^—^  to  forms  having  the  L.  C.  D. 

1— :c  1  — a;^  1—x^ 

The  L.  C.  D.  is  1  +  x—x^  —  x*. 

]  0.  Reduce r'  and  7 ^  to  forms  having  the  L.  C.  D. 

a^—b''  {a—b}'^ 

11.  Reduce ?  -^ ? ^to  forms  having  the  L.C.D. 

m-\-7i   m^+n^  m  +  n 

9        Q  2.7; 3 

12.  Reduce  — ?  ^ -?  'and ;  to  forms  having  the  L.C.D. 

x   2x—l  4^-2 _i 


lOo,  J^vob.  o.  To  Eeduce  Complex  Fractions  to  the 
form  of  Simple  Fractions. 

R  ULE. MULTIPIiY  NUMERATOK  AND  DENOMINATOR  OF  THE  COM- 
PLEX FRACTION  BY  THE  PRODUCT  OF  Alil.  THE  DENOjMINATORS  OF 
THE  PARTIAL  FRACTIONS  FOUND  IN  THEM  :  OR,  MULTIPLY  BY  THE 
L.   C.   M.  OF.  THE  DENOanNATORS  OF  THE  PARTIAL  FRACTIONS.* 

Dem. — TMs  process  removes  the  partial  denominators,  since  each 
fraction  is  multiplied  by  its  own  denominator,  at  least,  and  this  is 
done  by  dropping  the  denominator.  It  does  not  alter  the  value  of  the 
fraction,  since  it  is  multiplying  di%'idend  and  divisor  by  the  same  quan- 
tity. 

*  The  pupil  is  supposed  to  have  obtained  suflacient  knowledge  of  fractions  in 
-ionimon  arithmetic  to  perform  these  operations. 


116  FRACTIONS. 

EXAMPLES. 

1.  Beduce  —  to  a  simple  fractional  form. 


5a2 

MODEL   SOLUTION. 

2x 

OPEBATION.     = 

%  X  5.'  X  36^          ^^^,^ 

Explanation.  — In  order  to  free  tlie  numerator  of  its  denominator, 
2ic 
36*,  I  multiply  the  numerator  -—;  by  36'^  ;  but,  in  order  tliat  this  may 
ob 

not  change  the  value  of  the  fraction,  I  also  multiply  the  denominator 

4w 
by  the  same.     In  like  manner  to  free  the  denominator  -^    of   its   de- 
nominator, I  multiply  it  by  5a*  ;  but,  in  order  that  this  may  not  change 
the  value  of  the  fraction,  I  also  multiply  the  numerator  by  the  same. 

2r 

|^X5a*  X36*^ 

Indicating  these  operations  I  have .       To    multiply 

|,X5a^X36^ 

2x_ 

36^ 

ba^  gives  for  the  new  numerator  lOa^x.     So,   also,  I  obtain  the  new 

denominator  by  dropping  5a*  and  multiplying  hj  by  36*,  getting  there- 

10a*  X 
by  126*  2/.     Therefore  the  simple  fraction  is        ^     which    reduced    to 


QC 

to  a  simple  form.  Result,  — 


5-^  by  36*  I  drop  its  denominator  and  have  2x,  which  multiplied  by 


low( 

.    5a*a; 

jst  terms  is  77^5-. 

66*?/ 

2. 

Beduce 

m 

3. 

Beduce 

«+' 

4. 

Beduce 

cm 
cm 


m 


to  a  simple  form.  Result,  —    ,    ^» 


acn  -\-  hn 
to  a  simple  form.     Result,  ^,^^^ ^^^- 


ADDITION.  117 


5.  Reduce    — '■ —  to  a  simple  form. 

y 

5  3(1 

6.  Reduce  — -r \ —  to  a  simple  form. 

10  +  -i^o; 

a       X 

b         Tj 

7.  Reduce to  a  simple  form. 

a       X 

m      n 

b  —  a 


a  + 


8.  Reduce — to  a  simple  form. 

ab  —  a2 


SECTION  III, 
Addition. 

166,  I^rob,  To  add  Fractions. 

RULE. — Reduce  THEM  to  a  common  denominatok,  if  they 

HAVE  not    such   A  FOEM,    AND   THEN   ADD   THE   NUMEKATOKS,    AND 
WKITE   THE   SUM   OVEK  THE   COMMON   DENOMINATOE. 

Dem. — The  reduction  of  tlie  several  fractions  to  a  common  denom- 
inator, if  they  have  not  one,  does  not  alter  their  values  (163),  and 
hence  does  not  alter  the  sum.  Then,  when  they  have  a  common  de- 
nominator (divisor),  the  sum  of  the  several  quotients  is  equal  to  the 
quotient  of  the  sum  of  the  several  dividends  divided  by  the  common 
divisor,  or  denominator  {103). 

EXAMPLES. 
«„,.,,  ^1  -{-    X   1  -\-  X^  ,1  +  0^3 

1.  What  is  the  sum  of  _ , ,  and ■ . 

1  —  X  1  —  x^  1  —  x^ 


118  FRACTIONS. 

MODEL   SOLUTION. 

Opeeation.— The  L.  C.  M.  of  1  —  x,  1  —  x'^  and  1  —  x^  is  (1  —  x^) 
X  (1  +  x)  =  1  +  X  —  x3  _  a;4. 

(1  4-  a;)  X  (1  +  2x  +  2x^  +  x^)  _  1  -f  3x  4-  4x2  +  3x3  4.  3.4 
(1  —  X)  X  (1  4-  2ic  -f  2x^  +  x3)  ~  1  -j-  X  —  x^  —  X'      ~ 

(1  -{-  X-')  X  (1  +  X  +  x2)  _  1  +  X  -f  2x3  -f  x3  -)-  x^ 
(1  —  x2)  X  (1  +  X  +  x^)  ~  Hf  X  —  x^  —  X"*   • 

(1  +  x3)  X(l  4-  x)         1  4-  X  +  x"  +  x^ 
(1— x3)  X  (1  -\-x)   "^  1-l-x  — x3  — X4' 
1  .f  X   ,   1  4-  a52  ,   1  +  x3  _  1  -I-  3x  4-  4x2  -1-3x3  +x4 


X^  1  —  X3  1  4  *  —  ^^  —  ^'^ 

1  4-x 
3  -j_  5a;  4.  6x2  _|_  5a;3  _j_  3a;4 


1  -f-  X  4  2x2  4-  x3  4  x^       1  _}_  a;  4-  x3  -f-  x-« 
^         1  4-  X  —  X*  —  x^        ''  1  -|-  X  —  x3  —  x-»^ 


1  4"  ic  —  X-'  —  X* 

Explanation. — Explain  the  reduction  to  a  common  denominator  as 
under  {163)  unless  that  is  already  sufficiently  familiar. 

Having  reduced  the  fractions  to  the  L.  C.  D.  I  find  (read  A).     Now 

since  the  sum  of  these  quotients  is  equal  to  the  quotient  of  the  sum  of 

the  several  dividends,  or  numerators,  divided  by  the  common  divisor, 

or  denominator,   {103)  I  add  the   numerators   and  write   the  sum 

,,  _  .,....'      345x46x'45x3+3ic* 

over  the  common  denominator,  which  gives  — —- — r ^ h " 

"^  1  -f-  X  —  X-*  —  X* 

for  the  sum  of  - — ■ — ,  - — ■ — 7,  and  - — ' — — . 
1  —  xl  —  x^  1  —  X"* 

^.-_a7c  —  a       .  c  -\-  a  „        6^  4  7c  —  a 

2.  Add  — ,  -— — ,  and  -j — .  Sum,  — . 

a    c    6  a  a^h^  ch  +  h^  -^  a'^ 

3.  Add  -,  — r,  -,  and  -r-.  Sum, ■ — r . 

h  ah  a  ¥  a¥ 

^•^^^3'i'12'i8'6'"^^9-  ^^^''^• 

^.^^fl  —  3        ,54a  ^        3a  —  1 

5.  Add  — - —  and  — - — .  Sum,  — - — . 

6.  Add , ,  and Sum, 


14  a'  1  —  a'  1+a  '1  —  a 


ADDITION.  1 19 

7.  Add and .  Sum,  -— . 

1  —  a*           1  +  a^  .          '  1  —  a2 

8.  Add—— —  and .  Sum, 


1  +x  1  —  x  '  1  —  X-' 

9.  Add and 


Sum, 


as  —  a"-b  —  ab-^  +  63* 
10.  Add  — ^— ,  -4-  and  — i— . 

Sum,  ■ — — —^. 

x-^  —  y^ 

107 »  Cor. — Expressions  in  the  mixed  form  may  either  be 
reduced  to  the  improper  form  and  then  added,  or  the  integral 
parts  may  be  added  into  one  sum,  and  the  fractional  into  an- 
other, and  these  results  added. 

J. 2  3jc  4-  4 

1.  Add  Ix  +  — — -  and  Hx  -\ . 

O  uX 

FIRST    FORM    OF   OPEBATION. 
iC  —  2         22x  —  2 


7x 


3a;  4- 4  _  40x-^  4- 3x  +  4 
'   5x         5x 
(22r  —  2)  X  5a;  _  110  x'^  —  lOx 
3   X  5x  ~~    lox 
(40a;2  -I-  3x  4-  4)  X  3  _  120x^  -\-'dx-J^n 
5x      X^  ~~  15x 

110x2  _  lOx   120x-^  -f  9x  +  12  _  230xg  —  x  +  12 
15x  15x  15x 

SECONT)   FORM   OF   OPERATION. 

7x  4-  8x  =  15x 

(x  —  2)  X  5x  _  5x2  _  lOa; 

3    X  5x  ~    15x 

(*3x-|-4>  X  3  _  9x  +  12 

5x   X~3  ~   15x 


120  FRACTIONS. 

9^+12  ,  ,  5^2  _  j;  4.  12 

+  -^-  =  1'^  + — ^x — • 

EXPLA.NATION. — Since  the  sum  of  several  numbers  is  the  same  in 

whatever  order  their  parts  are  added,  I  take  the  integral  parts  first. 

Adding  Ix  and  8ic  I  have  locc.     Reducing  the  fractions  to  a  common 

.     ,       X  — 2,  5^2  —  lOx        ,  3x  +  4,  9ic4-12 

denominator,  — ~ —  becomes  — -— ,  and  — ~ — becomes  —-1 . 

3  15x  5a;  15x 

5.1-2  .T  4-  12 

Adding  these  I  have  — which  added  to  15x,  the  sum  of 

15x 

5a;2  x  4- 12 

the  integral  parts,  gives  for  the  entire  sum  15x  -| --z — — — . 

Idx 

2.  Add  ^Xy  3^  -}-  — -,  and  x  -\-  -^.        Sum,  6x  +  — — . 

5  9  45 

«     .  -,  -.  ^^-  .     .        2a^ 

3.  Add  a ^-  to  5  + . 

^                     ^  ^               ,       2abx  —  3cj;2 
bum,  a  4-  0  -f   7 . 

DC 

4.  Add  Ix  +  - — - — ,  and  dx . 

6  ox 

Sum,  Wx  +  - 


15a; 


5.  Add  6a; —,  — 8a;,  and  3a;  —  -. 


Sum,x-  --. 

«»i-i«        •      0,  4-  b         ^  a  —  b        -  . 

6.  Add  3ma; -,  and  7  —  2ma;  +  4. 

a  —  0  a  +  6 

Sum,  4  +  mx 


fl2 


7.  Add  6x^y^  —  3a; ^-r-^,  and    5a;   —   2x%^'   -f 

x-i  —  yi  Sum,  407^7/^  -)-  2a;  + 


07^  ^■^ 

r 

pq    '     pr    '  qr 


8.  Add  ^^— ^,  "^ ^,  and  ^^ -.  Sum,  0. 


SUBTBACTION.  121 

05          3a           -           lax  '  4a 

9.  Add , ,  and .  Sum, 


a  —  X  a^x  a^  —  x'^  a^x 

-„.,,         a^h                 6  +  c  c  +  a 

10.  Add  — -, -7 —,  and 


(6— c)(c— a)'  (c— a)(a— 6)  (a — b){b — c) 

Sum,  0. 

SuG. — The  L.  C.  D.  is  {a  —  h)(b  —  c){c  —  a),  since  this  contains  all 
the  factors  of  each  denominator,  and  no  more.  The  terms  of  the  1st 
fraction  must  be  multipUed  by  a  —  6,  of  the  2nd,  by  6  —  c,  and  of  the 
3rd,  by  c  —  a. 


I 


SUCTION  lY. 

Subtraction.  ^""^^^^ 

168,  JPvoh*  To  subtract  fractions, 

RULE. — KhDUCE  THE  FRACTIONS  TO  A  COMMON  DENOMINATOR, 
IP  THEY  HAVE  NOT  THAT  FORM,  AND  SUBTRACT  THE  NUMERATOR  OP 
THE  SUBTRAHEND  FROM  THE  NUMERATOR  OF  THE  MINUEND,  AND 
PLACE  THE  REMAINDER  OVER  THE  COMMON  DENOMINATOR. 

Dem. — The  value  of  the  fractions  not  being  altered  by  reducing  them 
to  a  common  denominator,  their  difference  is  not  altered.  After  this 
reduction,  we  have  the  difference  of  two  quotients  arising  from  divid- 
ing two  numbers  (the  numerators)  by  the  same  divisor  (the  common 
denominator).  But  this  is  the  same  as  the  quotient  arising  from  divid- 
ing the  difference  between  the  numbers  by  the  common  divisor(i04). 

EXAMPLES. 

1.    From subtract '-. 

X  —  y  X  -\-  y 

MODEL   SOLUTION. 
OPEKATION.       {X  —  y){x  -\-  y)  ^  X^  —  T/2 


(a;  4-  ?/)  X  (x  +  y) 

x'-  4-  2xy  +  y' 

{x  —  y}X  cx  +  y) 
(x  —  y)  X  (X  —  y) 

x:^  —  y^ 
x2  _  2xy  +  it/2 

(X  4-  y)  X  (X  —  y) 
-  {x'  -  2xy  -f-  y^)  = 
x-\-y      x  —  y 

x:'  —y^ 
=  ^xy 
4.xy 

(x'  4  2xy  +  y') 

x  —  y     x-\-y 


122  FRACTIONS. 

Explanation. — The  L.  C.  M.  of  x  —  y  and  x  -{-y  is,  tlaeir  product, 
since  they  have  no  common  factor.    Hence  x^  —  y^  is  the  L.  C.  D.     To 

X     I     XI 

reduce   -■      '   to  this  denominator  I  multiply  both  its  terms  by  a;  -j-  ^Z' 

3.2  _j_  2a;w  -4-  w* 

which  gives ^ —     2       •     ^^  ^i^®  manner  multiplying  both  terms 

X        y 

2« 71  0*2  2icw  -4—  ?y^  '' 

of  — — ^  by  X  —  y,  I  have — '-^ — .      I  have  now  to    subtract 

X  -j-  2/  x'  —y' 

±ZpLplf,om^^±p-pl.      Since    the   difference    of   the 

^  —  y  ^  —  y 

quotients  of  two  numbers  divided  by  the  same  number,  is  the  same  as 
the  quotient  arising  from  dividing  the  difference  between  those  num- 
bers by  the  common  divisor,  I  take  the  difference  of  the  numerators 
(the  quantities  to  be  divided)  which  is  ixy,  and  dividing  it  by  x*  —  y^ 

I  have  — — ^-^,  for  the  remainder  of— ^i^  less  — — ^. 
x^  —  y^  x  —  y  x-^y 

__  1+^^,1    X  „  .,  4:X 

2.  i  rom take .  Eemainder, 

1  — X  1  -\-  X 

3.  From    take .         Remainder, 

a  —  X  a   -\-  X 

X  ^  _i_  3 

4.  From   — -  take .  Remainder, 

X  —  6  X 

5.  From   3^  take '-.  Remainder, 

5  o 

3x 
Stjg. — Regard  3x  as  — . 

6.  From  9v  take  — - — -.  Remainder,  — =— . 

^8  8 

7.  From  r  take — -.  Rem..  — ; ■; . 

a  —  b  a' — 2a6  +  62  '       {a— by 

0-1:1  1.1        2  —  ^2  _  3^ 

8.  From take 


'1 

—  x^ 

2x 

a2 

—  x^ 

9 

x;^  ■ 

—  3^' 

3x 

—  3a 

X  +  4:  x^  -j-lOx  -\-  24' 

(^4-  2)2 


Remainder, 


x^  +  10^7  +  24  * 

^    ^          3x-{-2  ^  .     lax—  10a 
9.    From  ■ —  take . 

^                       ^'                            4(3  —  x) 
Remainder,  — -. 


SUBTRACTION.  123 

10.  Combine  the  foUowmsf  fractions  -:: f- 

^  2  —  ^  2  +  a;     ^ 

16x  —  x^  ^      ,         1 

Besult, 


x^  —  4:  '  x-\-2'^ 

11.  Combine   — 7—; +  -— }- 

a{a—  b){a—  c)    ^  h{b  —   c){h  —   a)  ^ 

— -, —.  Result,  — -. 

c{c  —  a){c  —  h)  abc 

^-    ^      ,  .        3a  —  4&        2a  —  h  —  c        15a  —  4c 

12.  Combine +  ^^ 

a  — 46  ^      ,^    81a  — 4^ 

BesulL 


21  ^*^"^^'''         84 

13.    Combine + ; . 

a  -\-  h        a-'  —  62        a:^  -\-  b'i 

Result, 


14.  Combine 


15.  Combine 


a*  —  h* 
3  7  4  —  2007 


1  —2x        l-^r  2x        4x^—1 

Result,  0. 
5 1  24 

2{x  +  1)         10(a;—  1)  ~  5(2;^  +  3)* 

2^ 3 

ResuU, 


'  (^2  _  l)(2;r  +  3)' 

16.  Combine \- ^ . 

x^—yi         {x-^yY         {x~yY 

_      .    x-2  —  4^V  —  y2 

ResuU,  -— ~. 

{x^  —  y-^Y 

169,  CoR. — Mixed  numbers  may  be  subtracted  by  annex- 
ing the  subtrahend  with  its  signs  changed,  to  the  Tninuend,  and 
then  combining  the  terms  as  much  as  may  be  desired.  The 
reason  for  the  change  of  signs  is  the  same  as  in  whole  numbers. 

{77.) 

EXAMPLES. 

17.  From  3x  +  ^±^  take  x  —  ^LZlL., 

^                          ^                             2ar 
Remainder f  2x  H . 


124  FRACTIONS. 

-lo-ni  4a  —  b       .  ,       ,   „  3a  —  2b 

18.  From  x —  subtract  7x . 

2  S 

Bemainder,  —    ,  —  6^  —  a. 
b 

-  „    Ti         o    X  1     3j:  +  12a  ^        .    ,      3a  —  3^ 

19.  From  Sa  take .  Bemainder, 

5  5 

20.  From  2x  +  ^^=1^  take  3x  —  "^fL+l, 

^                          ^   ^   .    ,      16^  +  23 
liemainder, . 


I 


SUCTION   V, 


k 


Multiplication, 

170,  I^voh*  1,  To  multiply  afi^action  by  an  integer. 

B  VLE. — Multiply  the  numerator  or  divide  the  denomi- 
nator. 

Dem. — Since  numerator  is  dividend  and  denominator  divisor,   and 
the  value  of  the  fraction  is  the  quotient,  this  rule  is  a  direct  conse- 
quence of  {101,  102).  \ 
EXAMPLES. 

.   ,     m  —  n  - 
1.  Multiply  — by  m  +  n. 


MODEL  SOLIJTIOW. 

m  —  n        ,       ,      ^      m 


OPEBATION.         — X  {m  -\-  n) 


dxy  dxy 

Explanation. — Since  m  —  n  is  divided  by  Sxy  if  I  multiply  it  by  / 
m  -{-n  9,nd  then  divide  (or  indicate  the  division),  I  shall  have  m  -\yn 
times  as  large  a  quotient  as  at  first.     But  the  value  of  a  fraction  is  ifne 
quotient  of  the  numerator  divided  by  the  denominator.     Hence  Bhul- 

»YI  /vy  771 71 '^  / 

tiplying  the  numerator  of  — by  m  -f  n,  I  have  — wh^ch  is 

m  —  n  j 

m  -\-n  times  — ;r .  -  ' 


MULTIPLICATION.  12i 


2.  Multiply  ^^  by  3a. 


MODEL   SOLUTION. 

2mx        „  2ma; 

Explanation. — Since  2mx  is  to  be  divided  by  Sa^b^  if  I  divide  the 
divisor  by  3a,  thus  making  it  3a  times  as  small,  it  will  go  into  the 

dividend  3a  times  as  many  times  as  before.     Hence  — r-^  is  3a  times 

2mx  , 


Za'h^ 


3.  Multiply Y  hj  X  -{-  y. 

dC 

4  Multiply   ^^^  _^^  hjx  —  y. 

5.  Multiply by  a-  +  1.  Prod.,   . 

Suggestion,     a^  —  a  z=  a{a*  —  1)  =  a{a^  —  !)(«'*  +  !)• 

6.  Multiply  — ^-±i^-  by  (a  +  yy. 

Suggestions. — Multiply  by  one  of  the  factors  of  (a  -\-yY  byreject- 

(a  -f-  VY 
immator,  gi\T.ng ■ — - — 


ing  it  from  the  denominator,  giving  ^^^  ^^      ^   ,  and  this  product  by  the 
other  factor,  giving 


am  — my- 

3 

7.  Multiply by  a^  —  x^  Prod.,  3a  +  3^. 

a  —  X 

8.  Multiply  -^^^  by  2a  +  2y.  Prod.,  ^^'. 

9.  Multiply  5^—^  by  36.  Prod.,  3a  —  2y. 

ScH. — A  fraction  is  multipHed  by  its  own  denominator  by  removing 

3a— 2,v 


it     In  the  last  example  it  is  evident  that  3a  —  2?/  is  36  times 

3a^ 
10.  Multiply by  a:  —  y. 


3b 


126  FRACTIONS. 

11.  Multiply  — ^—  by  2ax{x  —  y) .  Prod.,  Qa^x"^. 

X        y 

3c 

12.  Multiply by  x'^  —  1. 

X  —  X 

13.  Multiply^       ^  by  x^  —  2xy  +  y\ 

X      y 


17 1»  J*TOb.  2,   To  multiply  by  a  fraction. 

RULE. — ^Multiply  by  the  numerator  and  divide  by  the 

DENOMINATOR.* 

Dem. — Let  it  be  required  to  multiply  m,  which  is  either  an  integer 
or  a  fraction,  by  — . 

1st.  Suppose  a  and  h  are  both  integers.  Multiplying  m  by  a  gives  a 
product  6  times  too  large,  since  we  were  to  multiply  by  only  a  6th 

part  of  a  ;  hence  we  divide  the  product,  am,  by  h,  and  have  -r-. 

2nd.  When  either  a  or  h,  or  both,  are  fractions.  Let  c  be  the  fac- 
tor by  which  numerator  and  denominator  of  -  must  be  multiplied  to 

make  —  a  simple  fraction  {165).  Then  will  —  be  a  simple  fraction, 
i.  e.,  ac  and  he  are  each  integral ;  and  the  multiplication  is  effected  as 
in  Case  1st,  giving  -yr-     "^^^^  reduced  by  dividing  both  terms  by  c 

gives  — .  Hence  we  see  that  in  any  case,  to  multiply  by  a  fraction, 
we  have  only  to  multiply  the  multipHcand  by  the  numerator  of  the 
multiplier,  and  divide  this  product  by  the  denominator.  It  is  also  to 
be  observed  that  this  reasoning  apphes  equally  well  whether  the  mul- 
tiplicand is  integral  or  fractional. 

EXAMPLES. 

1.  Multiply  -^ by 7. 

__ _^ . — ^ —  ■  I 

*  It  is  assumed  that  the  pupil  knows  how  to  divide  a  fraction  by  an  integer,  from 
his  study  of  arithmetic.  Nevertheless  the  problem  will  be  introduced  hereafter  for 
the  purpose  of  familiarizing  the  pupil  with  the  literal  operations. 


MULTIPLICATION.  127 


MODEL   SOLUTION. 


OPEEATION.  -^ X  (X  —  y) 


a;3  — y3  -^         x^-j-xy+y 

.        (0—6)2  a_^ 

Again,—— ■ — -  —  (a  —  6) : 


^^-\-^y-^y''   '  ic^  +  a-t/  +  y2- 

(a  —  6)2     X — y a  —  6 


x3  —  y3      a  —  6       X-  -j-  xy  -f-  y^ 


Explanation. — In  order  to  multiply  — ; 7-  hx  ' f,  Ifirstimil° 

^  -"  x-^  —y^     "^  a~b 

tiply  by  X  —  y.  This  is  effected  by  dividing  the  denominator,  x^  —  y^, 
hjx  —  y,  {102),  and  gives  ^  — — - — j.  But,  since  this  multipHer 
was  to  be  divided  by  a  —  6,  the  product  now  obtained  must  be  di- 
vided by  the  same.     Dividing  —, — ; ■ — -  by  a  —  6  by  dividing  the 

^  x'  -\-xy  +  y^    ^  •"  ^ 

numerator  {101),  I  have  for  the  complete  product    ^ — - — j. 

x  -f-  ^y  -{-  y 

2.  Multiply—    by-.  Prod.,  — . 

«    ,.,..,      2a2  11^3^  22^7-2 

3.  Multiply  .— —  by  — --.  Prod.,  . 

4.  Multiply  -  -y  by  3-  .  Prod.,  -  —. 

5.  Mulhply  -  3^-  by  -  y,-.  Prorf..  ^^. 

ScH, — When  there  are  no  common  factors  in  the  numerators  and 
denominators  of  the  fractions  to  be  multipUed  together,  the  process 
consists  simplj^  in  multiplying  numerators  together  for  a  new  nume- 
rator and  denominators  for  a  new  denominator,  which  process  is 
equivalent  to  multiphdng  hy  the  numerator  of  the  multipHer  and  divid- 
ing by  the  denominator.  As  it  is  immaterial,  as  far  as  the  result  is 
concerned,  which  of  these  operations  is  performed  first,  that  one 
should  be  which  is  most  convenient  when  there  is  any  choice. 

2771 

6.  Multiply  4^2  —  9^2  by 


2j7  — 3?/ 


128  FRACTIONS. 

Suggestion. — In  this  case  divide  first,  obtaining  2x  -\-  dy.     Multi' 
ply  this  by  2m. 


7.  Multiply- by ^^. 

8.  Multiply   —- by       ^ 


Prod.  J  4:mx  +  Gm^/. 

Proc?. 

x^  -\~  ax 
■'   a2  +  ac 

ProtZ., 

X*  —  b* 
b^'C  +  bc<^ ' 

Frod., 

a'^  -f  6^ 

/                  .          7    \    -• 

6c  6  +  c 

9.    Multiply  ^  by — ; — -.  ^.^^.,      .      ,   ,.  . 

SuG. — In  the  last  example  both  operations  are  performed  upon  the 
denominator  of  the  multipUcand. 

,«-,,.,      3;r2  —  5x  ^           la  „     ,    3a,'r  —  5a 

10.   Multiply  -^^-  by  ^^^-^.  Prod.  j;^^-. 

11.  Multiply-^  "^yai^;;)-  ^'•"'■'i8- 

12.  Multiply   -^^  by  ^-^.  Prod., ——^ 

13.  Multiply  1^  by  1^.  Prod,  il^. 
14   Multiply  -^  by  — .  Prod.,  — — „. 

15.  MiUtiply  -  by  '-.  Prod.,  — -„. 

2/         **'  y 


16.   Multiply 


«*- 

-  >y 

-6^ 

-bi' 

Prod., 

ai 

+  hi 

«*- 

J 

-bi' 

3c  ~^x^           ^c^x 
17.    Multiply by -^   and   write    the    result 

5a^y~'^       la-^y~^ 

_,     ,     Qicx*y 
without  negative  exponents.  Frod.,   -or~7- 


MULTIPLICATION.  129 

18.   Multiply  together, — ,  —    (^  +  vY     ^^^    ___ 

Sm  _      ,  3m 

Prod., 


19.  Multiply ,^  ■  by  — —- — ^~-. 

Prod., ■ — . 

X  +  '6 

172*  Cor. — To  multiply  mixed  numbers,  Jii^st  reduce  them 
to  improper  fractions. 

20.  Multiply   1  —  ^  by  -  —  2. 

^     ,    Ixy  —  2^2  —  Qy2      x(7y—2x) 

Prod.,  —^ ^  or  -1-^.^— -Z_  2. 

3y-  Sy- 

21.  Multiply  together  —  yipy.  ^  ~  ^">  and  1  ^^   ^^^. 

?/-  1 


Prod., 


X 


^2  d  X  (V*  X* 

22.  Multiply  a  -  -  by  -  +  ^.               Prod.,  ^^. 

23.  Multiply  1   -   ^-^by2+^-. 

Prod,  ^"'^ 


jr2  _  ^2 
24.  Multiply  a;^  _  j;  +  1  by  —  +  -  +  1. 

Prod.,  ^=  +  1  +  — . 


OPEIUlTION. 


072   ^   +    1 

X        x^ 


J7  —  1+    - 
X 

x^  —  ,r  4-  1 

x^  +1  H ,  Product. 


130  TRACTIONS. 


SECTION  VL 
Division. 

173,   J*vob,  1,  To  divide  afrmtion  hy  an  integer. 

RULE. — Divide  the  numebator  or  multiply  the  de- 
nominator. 

Dem.  — Since  numerator  is  dividend,  and  denominator  divisor,  and 
the  value  of  the  fraction  the  quotient,  this  rule  is  a  direct  consequence 
oi  {101, 102,) 

examples. 

1.  Divide  -r r-  by  a  —  x. 

zmx  —  1 

Solution. — Since  the  value  of  this  fraction  is  the  quotient  of 
3(a*  —  a*)  divided  by  2mx  —  1,  if  I  divide  the  dividend,  3(a2  —  a^), 

by  a  —  X,  I  divide  the  fraction.     Hence — •  -^   (a  —   a)  = 

3a  4-  3.x 
2mx  —  1* 

2.  Divide  ^"^  ~"  t^^  by  a  +  6. 

a —  6 

Solution. — Since  the  value  of  this  fraction  is  the  quotient  of  3m 
—  4xT/  divided  by  a  —  &,  if  I  multiply  the  divisor,  a  —  b,  hj  a  •■{-  h,  I 

divide  the  fraction.     Hence  — '■ ^  -^  (a  4-  b)  =  — 3 :^ 

a  —  6  a'*  —  6* 

3.  Divide  ?J^  by  3^. 

.    ^.  .,     15a2  —  15j;2  ,      ^,  ,        ^  3(a — x) 

4  Divide  by  5{a  +  x).      QuoL,    -. 

X  —  y  X  —  y 

x^  —  1 
5.  Divide hj  x  —  1. 

X-'  +  6    -^ 


DIVISION.  131 


6.  Divide  —  m  x  —  y. 

X  -{-y 

7.  Divide  —  by  ^  +  y. 

X  -i-y 

8.  Divide by  x^  +  4. 

x^  —  4 

9.  Divide  ~  hj  x^  —  yK 

X  +  y 


SuG.     (X*  —  V*)  =  (a;  4-  y)ix  —  y).     Divide  by  the  factor  x  —  yhy 
.  .       a*  +  x^y  4-  x^y^  4-  X2/3  ^_  ^4 

dividing  the  numerator,  giving  ^ — '—•      r^ow 

X  -\-  y 

divide  this  quotient  by  the  other  factor,  «  -|-  y,  by  multiplying  the  de- 
x^  4-  x^V  +  ^""V^  +  xv^  +  x^ 

VioTnTnfl.rn'r      onvrnc  ' 


nominator,   giving    '     ^x+y)^ 

10.  Divide    ^^y\  by  5  mn^ A  QuoL,  ^ 

^    J-    -4  ^D?W 

,  ^    ^.  . ,     a2  —  2a6  +  &2 

11.  Divide -r by  a"  —  ft'. 

12.  Divide  ?^g^by42(a-6)=.    ' 


174.  Proh.  2,  To  divide  by  a  fraction. 

RULE. — Divide    by    the    numerator  and   multiply   the 

QUOTIENT  BY  THE  DENOMINATOR.  Or,  WHAT  IS  THE  SAME  THING, 
INVERT  THE  TERMS  OF  THE  DIVISOR  AND  PROCEED  AS  IN  MUL- 
TIPLICATION. 

Dem. — The  correctness  of  the  first  process  appears  from  the  fact  that 
division  is  the  reverse  of  multiphcation,  and,  hence,  as  we  multiply 
by  the  numerator  and  divide  by  the  denominator  in  order  to  multiply 
by  a  fraction,  to  divide  by  one  we  must  divide  by  the  numerator  and) 
multiply  by  the  denominator. 

The  process  of  inverting  the  divisor  and  then  multiplying  by  it  is 
seen  to  be  the  same  as  the  other,  since  this  multipUes  the  dividend 
by  the  denominator  of  the  divisor  and  divides  by  the  numerator. 


132  FRACTIONS. 

Again,  this  process  may  be  demonstrated  thus :  Inverting  the  divisor 
shows  how  many  times  it  is  contained  in  1.  Then  if  the  given  divisor 
is  contained  so  many  times  in  1,  it  will  be  contained  in  5,  5  times  as 
many  times  ;  in  f ,  f  as  many  times  ;  in  ax^,  ax*  times  as  many  times ; 
or  in  any  dividend  as  many  times  the  number  of  times  it  is  contained 
in  1,  as  is  expressed  by  that  dividend,  whether  it  be  integral,  fractional 
or  mixed.     (The  author  prefers  this  demonstration.) 

ScH.  1. — Since  to  multiply  one  fraction  by  another  we  may  multiply 
the  numerators  together  for  the  numerator  and  the  denominators  for 
the  denominator,  and  since  division  is  the  reverse,  we  may  perform 
division  by  dividing  the  numerator  of  the  dividend  by  the  numerator 
of  the  divisor,  and  the  denominator  of  the  dividend  by  the  denomina- 
tor of  the  divisor. 

This  method  will  coincide  with  the  others  when  they  are  worked  by 
performing  the  operations  by  division  as  far  as  practicable,  and  this  is 
worked  by  performing  the  multipHcations  equivalent  to  the  divisions 
when  the  latter  are  not  practicable. 


EXAMPLES. 


1.  Divide      ^^'      by  -^ 

a;3  -f  2/^        ^  + 


MODEL.  SOLUTIONS. 

Operation  by  the  First  Method. 


2yg  _      2y 

a.3  _j_  yi  —  y—^T^Ty 


3' 


a.3_f_y3  Xix  +  y)_  ^,  _^y_^y. 

Explanation. — I  first  divide  -— -^ — -  by  y,  by  dividing  the  numera- 

*    +  y 
tor  {101).     But  the  given  divisor  is  y  divided  by  a;  -f-  y  ;  and  as  divid- 
ing the  divisor  multiplies  the  quotient,   {102),  I  must  multiply  this 

2v 
quotient,  -j-~— j  by  x  4-  !/•     Performing  this  by  dividing  the  denom- 

X    -\-  y 


inator,  I  have  for  the  true  quotient  ^^  \    f 


III. 
X'  —xy-^y-' 


Operation  by  the  Second  Method. 
2y^         .       y      _      2.v2  x -^  y  2.y 


ic  -\-  y*  '  X  '{-y     x^  -f-  y^       y        ^^  —  ^ 


DIVISION.  133 

Explanation. — By  inverting  the  divisor  and  indicating  the  multipli- 

cation  of  the  dividend  by  it,  I  indicate  that  the  dividend  —t-~ is  to 

x-^  +  y^ 

be  multiplied  by  x  -[-  V  and  divided  by  y,  which  are  the  operations 

required.     In  this  instance  I  perform  the  multiphcation  hy  x  -{-  y,  by 

dividing   the    denominator,  and   the   division  by  y,  by  dividing  the 

numerator. 

The  operation  by  the  third  method  is  of  the  same  form  as  the  last. 

Explanation  by  the  Third  Method. 

I  am  to  find  how  many  times  — - —  is  contained  in ,      Now 

x  +  y  ic^  +  y' 

y    •         .      .      x  4-  V  .        .  1 

— ; —  is  contained  in  1,  '-  times,  since  y  goes  into  1,  —   times,  and 

^-hy  y  ^^  y 

—^, —  goes  X  4-  V  times  -,  or  times.     Hence  if  — - —  c:oes  into  1 

^+y  "^         y        y  x+y  ^ 

times,   it  goes  into  — -^ -,  —-^ times  — —^  times,  which 

y  x'  +  y'   X'  -{-y^  y 

2v 
I  find  by  the  rules  of  multiphcation  to  be . 

ic-  —  ^*/  +  y' 

2.  Divide by .  QuoL, -. 

(a  +  ^'Y'    '  a^  —  x^  ^         x{^a  +  x) 

3.  Divide  ^^-^  by  -5"A_  Q^ot,  2fa-by 

b{a  -\-  b)      *"  a-'  —  6^  -^02 

4  Divide ■ by  — - — ■ 

x-^  —  y^  -^  X-'  -^  xy  +  y^ 

Quot.,  ^±i:. 

x —  y 
5.  Divide  5^  by  ^.  ■  Quot.,^-^. 

a-'  —  zab  4-  b'^     ''    a  —  b  ^         a 


134 


FRACTIONS. 


9.  Divide  —-7 f  by --. 


10.  Divide  x 


2x 


hjx 


2x 


x  —  3  '*'  X  —  3' 

Suggestion. — Reduce  to  improper  fractions. 


QuoL, 


4(a  +  X) 
3(c  —  x)' 


QuoL, 


11.  Divide  x* 


X'* 


by  X 


SuG. — This    quotient     should    be   written    by    inspection   in    the 

same  manner  as  {x*  —  y^)  ^  {x  —  y),  and  is  x'"*  -f  x  -j j -.     Or 

it  may  be  performed  as  follows  : 


—  x2 


X  ■ 

X 


a;:i  _|_  X  +  i  +  L. 


1-^ 


12.  Divide 


1  -{-  X       1  —  X 


X         ,  1 

by:; 


1  +  X 


-.      QuoL,  1. 


l  +  x    '    1 

13.  Divide 


SUGGESTIONS. 

1  +  a-2  1 

— ^--  and  

1  —  ;^^ 


l_ic       1  4-x       1 


-^  + 


?by 


X  —  y       X  -\-  y        X  — 


X  +  y 


r,      ,     ^'  +  y^ 


DIVISION.  135 

-4  _  1  _  2  _  1 


30  tA/ 

14.  Divide  — — by 


15.  Divide '"^-'^%y^"- ^•^. 

^1^0^.,  3;r2  —  xy  —  2?/2. 

ScH.  2. — It  is  Bometimes  convenient  to  write  the  divisor  under  the 
dividend  in  the  form  of  a  complex  fraction,  and  then  reduce  the  result 
to  a  simple  fraction  by  {105). 

16.  Divide  1  +  -  by  1 -. 

a  a- 


OPEEATION. 

1  +  -1 X  a2 

a- 

■f-  a 

a 

1- 

In 

Xaa 

a2  J 

a' 

—  1 

a  —  1 

17. 

Divide  -  -\ — r 
a       ab^ 

by6  +  i-l. 

OPEEATION. 

rl 

-a 

ab^ 
ab^ 

?>3    +    1 

"6-f 

ab^ 

-^  ab'  —  ab'^ 

b+  1 

^3   +    1 


Suggestions.  


18.  Divide  x  ~^y  —  1/2  j^y  j??/  -2  -|-  j^  -2^. 

a;;/  -'  4.  x-'j/       rjc_  _^  f]  ^  ^.,,^,         X*  +  xy: 

19.  "Free  -r— - — ' from  nef2:ative  exponents. 

ab~^  -^  x~^  ° 

,     b^x^y^  —  b'^x 

Result,  ^ . 

axy^  +  b^y- 


136  FRACTIONS. 

20.  Free  —^^ -r—z:  from  negative  exponents. 

Itesult, 


21.  Free  — — - — r-  from  negative  exponents. 


Ans. 

22.  Free — ^  from  neefative  exponents. 

1  —  a:~^y-'-^-\-x~' 

*'      a-x-^y-  —  a--\-a^xy'^ 

%1CII 

23.  What  is  the  reciprocal  of  — ^'  '  ,  •  ? 

a'^  —  b- 

Solution. — The  reciprocal  of  a  quantity  being  the  quotient  of  1  di" 

vided  by  that  quantity,  the  reciprocal  of ——  is  1  -^ —,    oi 

a-  —  b- 

This  is  analyzed  thus  :   dividing  1  by  2xy  the  quotient  is  ^— .     But, 

dividing  the  divisor,  2xy,  by  a^  —  ?>-',  is  equivalent  to  multiplying  the 

2.TV  1 

quotient  :  hence  the  quotient  of  1  divided  by '--  is  - —  X  («"^  —  b-), 

^  ■  a- — b^      2xy 


2xy 

24.  AVhat  is  the  reciprocal  of  — j-^ — -  ?  Ans.,     ^'     . 

(a 5)-5 

25.  What  is  the  reciprocal  of  ■ •,  ? 

Ans.,   a-'  —  3a2?)  +  3a62  —  63. 

175*  ScH. — The  reciprocal  of  a  quantity  being  1  divided  by  that 

quantity,  the  reciprocal  of  a  fraction  is  the  fraction  inverted.     Thus 

, ,  .  1     ^  o  •>  •      1      .2a  .    ^x^      „  a—b  .    a-j-b      „   1     , 

the  reciprocal  of  3a';^  is  r— ,of  —  is  — ,  of is  — ■ — ,  of  — -  is 

dx'       ^x"'      2a         a-^b      a  —  b         2x 

2x,  etc.     {174,  Explanation  of  3rd  method). 


SYNOPSIS. 


137 


Synopsis  of  Fractions. 


O 

t 

C 


I 


CQ 


t5   J 


FracHo7i. 


i  Numerator. 
/  Denominator, 


{ Integral. 
Foi-ms.   <  Fractional, 
(  Mixed, 


'  Tei*ms. 
Value  of  fraction. 
Cor.  1.  mult,     or    div. 

num.  or  denom. 
Cor.  2.  Remov.  denom. 
(Proper,       /Simple, 
I  proper,  jg-P-'l. 


Lowest  tei-ms. — L.  C.  D. — Reduction. 

Of  numerator,  of  denominator,  of  fraction. 
Essential  sign  of  fraction. 

f  Prob.  1.   To   lowest    terms.     Rule.      Bein. — 
Sch.     By  H.  C.  D. 
Fi'ob.  2.  From  impr.  to  int'g.  or  mx'd  forms.  Rule. 

I)e)n.  —  Cor.     Neg.  exp't. 
Pi'ob.  3.  From    integ.    or   mixed   to   frac.    forms. 

Rule.     Bern. 
Prob.  4.   To  com.     denom.     Rule.       Bein. — 
Cor.     L.  CD.     Devi. 
[  Prob.  5.   Complex  to  simple.     Rule.     Devi. 

Add. — P?'ob.     Rule.     Dem. — Cor.     Mix'd  Nos. 
Subfn. — Prob.     Rule.     Devi. — Cor.      Mixed  Nos. 
f  Prob.  1.  Frac.  by  integer.     Rule.     Devi. 


§  1 


3Iult. 


Divis. 


Frac.  by  integer, 
Sch. 

P7'ob.  2.  Any  No.  by  Frac.     Rule.     De7n. 
Sch. — Cor.    Mix'd  Nos. 

1.  Frac.  by  integer.     Rule.     Devi. 

2.  Any  No.  by  Fraction.     Rule. 
1. 

Devis.   -12.  \-  Sch's.   •{  2. 


{Prob 
J  Prob 


III 


Test  Questions. — Upon  what  five  principles  in  Division  are  most  of 
the  operations  in  fractions  based  ?  Why  does  the  process  of  reducing 
to  a  common  denominator  not  change  the  value  of  a  fraction  ?  Give 
the  rules  for  Multiphcation  and  Division  of  Fractions,  and  the  reasons 
for  them. 


138  POWEKS   AND   ROOTS. 

CHAPTEE  IV. 

PO  WEUS  AND  HOOTS. 


SUCTION'  I. 
Involution. 

[Note. — The  subjects  treated  in  this  chapter  are  among  the  most 
difficult,  if  not  actually  the  most  difficult  for  the  pupil  in  the  whole 
science.  In  the  examination  of  hundreds  of  students  from  all  parts 
of  the  country,  the  author  has  found  that  the  rule  is  that  they  are 
deficient  hi  knowledge  of  Badlcals.  An  attempt  is  here  made  to 
assist  the  teacher  in  remedying  this  defect,  by  constantly  holding  the 
attention  to  the  one  central  principle,  that 

The  operations  in  Radicals  are  all  based  upon  the  most 
elementary  principles  op  factoring, 

Many  of  the  demonstrations  can  be  given  in  more  concise  and 
elegant  forms  ;  but  it  is  thought  better  to  hold  the  attention  to  the 
single  line  of  thought.  If  the  student  learns  how  to  use  this  key,  he 
can  unlock  all  the  mj^steries  of  the  subject. 

The  desire  to  impress  the  central  principle  has  led  to  several 
repetitions.  ] 

GENERAL  DEFINITIONS. 

176.  A  I*OW€ri^  i\'2^'f'oduct  arising  from  multiply- 
ing a  number  by  itself.  The  Degree  of  the  power  is 
indicated  by  the  number  of  factors  taken.  Thus  2,  4,  8, 
16,  and  32  are,  respectively,  the  1st,  2nd,  3d,  4th,  and  5th 
powers  of  2. 

ScH. — It  will  be  seen  that  a  power  is  a  species  of  composite  number 
in  which  the  component  factors  are  equal. 

177*  A.  Hoot  is  one  of  the  equal  factors  into  which 
a  number  is  conceived  to  be  resolved.  The  Degree  of 
the  root  is  indicated  by  the   number  of    required   factors. 


INVOLUTION.  139 

Thus,  2  is  the  1st  root  of  2,  the  2nd  root  of  4,  the  3rd 
root  of  8,  the  4th  root  of  16,  the  5th  root  of  32,  etc. 

178.  ScH.  1.— Poioer  and  J^oo^  are  correlative  terms.  Thus  32  is 
the  5th  power  of  2,  and  2  is  the  5th  root  of  32. 

179-  ScH.  2. — The  Second  Power  is  also  called  the  Square;  the 
Third  Power,  the  Cuhe ;  and,  sometimes,  the  Fourth  Power,  the 
Biquadrate.  In  like  manner  the  2nd  root  is  called  the  square  root ; 
the  third  root,  the  cube  root ;  the  fourth  root,  the  biquadi-ate  root. 
These  are  Geometrical  terms  which  have  been  transfen-ed  to  other 
branches  of  mathematics.  The  second  power  is  called  the  square, 
because,  if  a  number  represents  the  side  of  a  square,  its  second 
power  represents  its  area,  or  the  square  itself.  Conversely,  if  a 
number  represents  the  area  of  a  square,  the  square  root  represents 
the  side.  Also  the  third  power  represents  the  volume  of  a  cube,  the 
edge  of  which  is  the  first  power  or  cube  root.  Biquadrate  means 
twice  squared,  and  hence  the  fourth  power. . 

180.  An  Exx>onent  or  Index,  is  a  number 
written  a  Uttle  to  the  right  and  above  another  number, 
and  indicates 

1st.  If  a  Positive  Integer,  a  Power  of  the  number  ; 

2nd.  If  a  Positive  Fraction,  the  numerator  indicates  a 
Power,  and  the  denominator  a  Boot  of  the  number  ; 

3d.  -ZT  a  Negative  Integer  or  Fraction,  it  indicates  the 
Reciprocal  of  what  it  would  signify  if  positive. 

III.  4^  is  the  3rd  power  of  4,  or  64.  a"'  is  the  mth  power  of 
a,   if  m  is  an  integer.      4-^   (read  "  4,  exponent  —  3  ")  is  —  or  — • 

.1  3.  . 

a  ~  "    is    — .      S'^  is  the  cube  root  of  the  square,  or  the  square  of  the 

cube  root  of  8,  or  4.    a"  is  the  ?jith  power  of  the  nth  root  of   a,  or  the 

nth  root  of  the  mth  power,  if  m  and  n  are  both  integers.     a~ ";;  is  — • 

a"" 
ScH. — It  is  obviously  incorrect  to  read  4^,  "the    f   power    of  4." 

There  is  no  such  thing  as  a  2-fifths  power,  as  will  be  seen  by  consider- 
in 

ing  the  definition  of  a  power.     Kead  4*^,  "4  exponent!;"  also  a", 


140  POWEES  AND   HOOTS. 

m, 

"  a  exponent  -" ;  a~  "5^,  "a  exponent  —  ^ ."  These  are  abbreviated 
forms  for,  "a  with  an  exponent  —  "^j"  etc.  In  this  way  any  exponent, 
however  compHcated,  is  read  without  difficulty. 

181,  Cor. — A  factor  can  he  tmmf erred  from  numerator 
to  denominator  of  a  fraction,  or  vice  versa,  hy  changing  the 
sign  of  its  exponent,  without  altering  the  value  of  the  fraction. 

„,1      ^  a"* 

^^      a^x--        a^y^     „     a'^x,--  ""  x^' _a^  _  or_      V' _ 

Thus  - — -  =  -r^-  ;  lor  -7-^3-  =  — 7"  —  ";  ^»  ■^  a  — 

182,  A  Hadical  JS'iiniber  is  an  indicated  root 
of  a  number.  If  the  root  can  be  extracted  exactly,  the 
quantity  is  called  Rational ;  if  the  root  can  not  be  ex- 
tracted exactly,  the  expression  is  called  Irrational,  or  Surd. 
Thus  the  radical   v''25a"2  is  rational,  but    ^V6a  is  surd. 

18 S.  A  Boot  is  indicated  either  by  the  denominator 
of  a  fractional  exponent,  or  by  the  Hadiccil  SigUf  V- 
This  sign  used  alone  signifies  square  root.  Any  other 
root  is  indicated  by  writing  its  index  in  the  opening  of  the 
V  part  of  the  sign.  Thus  Vam,  Vam,  are  the  3rd  and 
5th  roots  of  am,  and  the  same  as  {am)^,  (am)'^. 

184.  An  Imaginary  Qwaii^fif^y  is  an  indicated 
even  root  of  a  negative  quantity,  and  is  so  called  because 
no  number,  in  the  ordinary  sense,  can  be  found,  which 
taken  an  even  number  of  times  as  a  factor,  produces  a 
negative  quantity.  Thus  V— 4  is  imaginary,  because  we 
can  not  find  any  factor,  in  the  ordinary  sense,  which 
multiplied  by  itself  once  produces  —  4.  Neither  +  2  nor 
—  2  produces  —  4  when  squared.  For  a  hke  reason 
V —  3a2,    v^  —  bx,  or  v^  —  IM^xy"^  are  imaginaries. 

18S'     AH  quantities  not  imaginary  are  called  Real. 


INVOLUTION.  141 

186.  Similar  Iladicals_a.re  like  roots  of  like 
quantities.  Thus  4v^5a,  3^^5a,  and  (a*  —  x^)^ba 
are  similar  radicals. 

187.  To  nationalise  an  expression  is  to  free  it 
fi-om  radicals. 

188.  To    affect    a    number  tvitJi  an  Ex- 

ponent  is  to  perform  upon  it  the  operations  indicated  by 
that  exponent.  Thus  to  affect  8  with  the  exj^onent  f  is 
to  extract  the  cube  root  of  the  square  of  8,  or  to  square 
its  cube  root,  and  gives  4. 

180.  Til  vol  lit  ion  is  the  process  of  raising  numbers 
to  required  powers. 

100,  JEvoltttion  is  the  process  of  extracting  roots 
of  numbers.  * 

101.  Calculus  of  Radicals  treats  of  the 
processes  of  reducing,  adding,  subtracting,  or  performing 
any  of  the  common  arithmetical  operations  upon  radical 
quantities. 


INYOLUTIOX. 

102.     JProh.     1.    To  raise  a  number  to  any   required 

power . 

RULE. — Multiply  the  number  by  itself  as  many  thies, 

LESS  ONE,  AS  THERE  ARE  UXITS  IN  THE  DEGREE  OF  THE  POWER. 
Dem. — Since  the  number  of  factors  taken  to  produce  a  power,  is 
equal  to  the  degree  of  the  power  {170),  it  follows  that  to  obtain  the 
2nd  power  we  take  two  factors,  or  multiply  the  number  by  itself  once ; 
to  obtain  the  3rd  power  we  take  three  factors,  or  multiply  the  number 
by  itself  tuoice ;  and  in  hke  manner  to  obtain  the  ?ith  power  we  take 
n  factors,  or  multiply  the  number  by  itself  n  —  1  times. 

EXAMPLES. 

1.     What  is  the  3rd  power  of  la'  ? 


142  POWERS  AND   ROOTS. 

Model  Solution. — Since  the  3rd  power  of  2a*  is  the  product  aris- 
ing from  taking  it  3  times  as  a  factor,  I  have  ^a^  X  2a^  X  Sa^  =  8a  ^. 

2.  What  is  the  4th  power  of  —  lOa^  ? 

(—  lOa^j    X  (—  lOa^)  X  (  —  lOtt^)   X  (—  lOa^) 

=  lOOOOa^.  Ans. 

3  to  6.  What  is  the  square  of  —  hm'^n  ?     Of  6a6  "  ^  ?     Of 

5  b¥ 

ff2      9        9 

Answers,  25m^n%  367—,  -r^^,  -T^a}'^h~*. 
0      25a-"  25 

7  to  10.  What  is  the  cube  of  —  2ab'-  ?    Of  12m~  ^  ?     Of 

_2  „,     of-^? 

S  on  • 

-3         1728         8  m6 

Ansivers,  —  8a''6«,  1728m    ^  =  -,  —  — -  x^", 


11.  What  is  the  square  of  2a  ■ —  3^? 

SuG.     (2a  —  3x)  X  (2a  —  3x)  =  ia^  —  12ax  -\-  9x«.     Ans. 

12.  What  is  the  3rd  power  of  2a-='+  3^? 

Ans.,  8a«  +  36a 'a;  +  54a-'^2  +  27 xK 
13  to  17.  Expand  the  foUowing :  (1  +  2^  +  3^2)2^ 

(1  —  a-  +  ^•-'  —  x^y,  {a  +  b  —  cy,  (1  +  2a;  +  072)3,  an.l 

(    1   —   3^'   +    3^2   _  j;3)2. 

BesuUs,  1  +  4^  +  10^2  _|_  12^3  _|_  9^4^  1  _  2a:  +  3^» 
—  4a:3  +  3x^  —  2a;^  +  x%  1  -{-  Gx  +  15x^  +  20a:3  +  15a7-« 
4-  Gx^  +  x%  and  1  —  6a;  +  15a;2  —  20a;3  +  15x^  —  6x' 

+  a;«. 

18.  Show  that  ^ — — +  -7-^^-7 ^- 

64a'6^  64a26'< 

=  62. 
•19.  What  is  the  square  of  9a;  -f  -  ?     Ans.,  81a?2  +  18+  — . 


INVOLUTION.  14.3 

20.  Expand  as  above  (4  —  x^y.    Result,  16  —  So;"^  +  x. 

21.  Expand  (a"^'  +  c^)\ 

Remit,  a^  +  ia^c"^  +  Qac  +  4a^c^  +  cK 

103*  Cor. — Since  any  number  of  positive  factors  gives  a 
positive  product,  all  p)owers  of  positive  monomials  are  positive. 
Again,  since  an  even  number  of  negative  factors  gives  a  posi- 
tive product,  and  an  odd  number  gives  a  negative  product,  it 
follows  that  even  powers  of  negative  numbers  are  positive,  and 
odd  powers  negative. 


194,  Froh,  2,  To  affect  a  monomial  with  any  exponent. 

R  CLE. — Perform  upon  the  coefficient  the  operations 
indicated  by  the  exponent,  and  multiply  the  exponents  of 
the  letters  by  the  given  exponent. 

Dem.— 1st.    Wlcen  the  exponent  by  which  the  monomial  is  to  he  affected 

n 

is  a  positive  integer.  Let  it  be  required  to  aflfect  4a'»&'"x— «  with  the 
exponent  p ;  or  in  other  words  raise  it  to  the  pth  power,  p  being  a 

n  n 

positive  integer.      The   pth    power  of   4a'"6'"a;  — *   is    4a"'6'x  — *    X 

n  n 

4a"»6''x— *  X  4a'"6'"«""* top  factors.     But  as  the  order  of  the 

arrangement  of  the  factors  does  not  aflfect  the  product  (83),  it  may 
be  considered  as,  p  factors  each  4,  into  p  factors  each  a"',  into  p  fac- 

n 

tors  each  6%  into  p  factors  each  x—\  Now  p  factors  each  4  give  4''  by 
definition,  p  factors  each  a"'  are  expressed  a  p"^,  since  a"*  is  m  factors 
each  a,  and  p  factors  containing  m  factors  each,  make  in  the  whole  pm 

n  pn  n 

factors,  or  aP".    Again  p  factors  each  b'  are  expressed  &  '• ,  since  b''  is  n 

factors  each  &',  and  p  factors,  containing  n  factors  each,  are  pn  factors 

i          ?!?                                      1                            111 
each  b\  or  &^     And  since  a;-*  =  -,  p  factors,  or  -   x  -  x to 

p  factors  make  — ,  as  fractions  are  multiplied  by  multiplying  nume- 
rators together  for  a  new  numerator  and  denominators  for  a  new  d*» 


144  POWERS  AND   ROOTS. 

nominator,  and  ic*  X  ic"  X  «* to  p  factors  are  x  p'.       But  -^  = 

x  —  PK     Hence  collecting  the  factors  we  find    that   {4:a^b^x~^)p  = 

pn 

Ap  ap'^b "  x -p^    Q.  E.  D. 

2nd.    When  the  exponent  is  a  positive  fraction.     Let  it  be  required  to 

--  p  — 

affect  4a'"  &'"  ic -^  with  the  exponent  -.     This  means  that  4a"' 6  *■  x-*' is 

q 

to  be  resolved  into  q  equal  factors  and  p  of  them  taken.     Now,   if  we 

n 

separate  each  of  the  factors  of  4a"'  &  '■  x  — *  -into  q  equal  factors,  and  then 
take  p  of  each  of  these,  we  shall  have  done  what  is  signified  by  the 

exponent  — . 

q  l_ 

By  definition,  4^    represents  one  of  the  q  equal  factors  of  4. 

To  obtain  one  of  the  q  equal  factors  of  a  "*,  we  take  one  of  the  q  equal 
factors  of  a  from  each  of  the  m  factors  represented.     But  one  of  the 

q  equal  factors  of  a  is  represented  by  a'^,  and  m  of  these  is  ai  by  defi- 
nition. 

n  n 

To  separate  h »'  into  q  equal  factors,  we  notice  that  b~  is  n  of  the  r 
equal  factors  of  h.  Now,  if  we  resolve  each  of  these  r  factors  into  q 
equal  factors,  b  is  resolved  into  rq  equal  factors  ;  doing  the  same  with 
each  of  the  n  factors  represented,  and  taking  one  from  each  set,  we 
have  b  resolved  into  i^q  equal  factors  and  n  of  them  taken  ;  that  is 

?>'-^is  071C  of  the  q  equal  factors  of  & '' . 

To  resolve  x  -*  =  —  into  q  equal  factors,  we  consider  that  a  fraction 

Xs  ^ 

is  resolved  by  resolving  its  numerator  and  denominator  separately. 
But  one  of  the  q  equal  factors  of  1  is  1  ;  and  one  of  the  q  equal  factors 

of  X*  is  X''    as  seen  in  the  resolution  of  a  ™.     Hence  one  of  the  q  equal 

1-1  -- 

factors  of  X  —*  or  -  IS  -  ^  a;    <  • 
X*         f 
XI 

Collecting  these  factors  we  find  that  one  of  the  q  equal  factors  of 


4a"'6'x-*is  4'(a''6'''-ar     ''•     And  finally  p  of  these  being  obtained  ac- 

p     pm   pn         j)S  n 

cording  to  Case  1st,  gives  4 '' a  ''  h'''  x     "^ ,  as  the  expression  for  Aa'-b~x~  - 


INYOLUTION.  145 


affected  with  the  exponent^  ;  which  rssult    agrees    with    the  enun- 

q 

ciation  of  the  rale. 

3rd.    When  the  exponent  is  negative  and  either  integral  or  fractional. 

n 

Let  it  be  required  to  affect  4a'»  b'x-'  with  the  exi^onent  —  t.  This, 
by  definition  of  negative  exponents,  signifies  that  we  are  to  take  the 
reciprocal  of  what  the  expression  would  be  if  t  were  positive.     But 

n  '" 

4a'"&''x~*  affected  with  the  exponent  t  (positive)  is  4:'a""7>  ^x~*^  by  the 
preceding  cases,  whether  t  is  integral  or  fractional.     The  reciprocal  of 

this  ig .     But  since  these  factors  can  be  transferred  to  the 

4'a'"'f)  »■  X  -'* 
numerator  by    changing    the    signs    of    their    exponents,    we   have 

fn  "_ 

4  -*a  —^"'b~  r  x*' ,  as  the  result  of  afi'ecting  4a"'  hr  x-'  with  the  expo- 
nent —  t,  which  result  agrees  with  the  enunciation  of  the  rule. 

[Note. — The  above  demonstration  contains  the  fundamental  prin- 
ciples of  the  whole  subject  of  the  Theory  of  Exponents,  and  it  is  of  the 
highest  importance  that  it  be  made  perfectly  famihar.  The  applica- 
tion of  the  rule  is  so  simple  in  practice  as  to  afford  no  difficultj^  but  in 
each  of  the  following  examples  the  reasoning  should  be  given  in  full. 
The  purpose  is  to  fix  in  the  mind  these  principles  of  the  theory.  After 
this  is  done,  of  course,  the  expert  merely  performs  the  operations. 
The  danger  is  that  the  how  being  so  simple,  the  why  will  be  disre- 
garded.] 

EXAMPLES. 

1.  Affect  2a25^c~*  with  the  exponent  5  :  that  is,  raise  it  to 
the  5th  power. 

MODEL   SOLUTION. 


2.  1  ( 

Operation.  {^la'^h^c  -  *y  =  ^la^^lf^ 


Explanation.      {la^h^c-'^Y   is  2a^ly'c-'>'  X  2a«6^c-^  X  'la^ly'c-''- 
to  5  factors.      This  gives  5  factors  of  2,  or  32  ;  5  factors  of  a^  or 

2. 

10  factors  of  a,  a^"  ;  5  factors  of  6 •^   which  I  obtain  by  considering 

2.  -L  2 

that  h^  is  2  factors  of  h'\  and  hence  5  factors  of  h^  is  5  times  2  factors 
of  &•*  or  6  ^~  ;  and  since  c~*  =  — r-.  5  factors  of  it  aje  — -  X  — r  -  -  to 


146  POWERS   AND   ROOTS. 

5   factors  = -7-7  =  c- 20,    as    fractions   are   multiplied  by  multiplying 

2. 

numerators    and    denominators.        Hence    I    have    (2a^b'^c~*)^  =i 

LQ. 

<2 

2.  Affect  3a^  with  the  exponent  m  ;  that  is  raise  it  to  the 


mth  power,  m  being  an  integer. 


3.  2.  3.  3. 

Explanation.     (3a^)"'  =  3a=^  x3a^X3a^ to  m  factors.     This 

3. 

gives  m  factors  of  3  and  m  factors  of  a^.     But  m  factors  of  3  are  rep- 

resented  by  3™.     To  obtain  m  factors  of  a\  I  consider  that  a    is  2 

J-  "        ^  . 

factors  of  a^  ;  hence  m  factors  of  it  are  2m  factors  of  a**.     That  is  a 

resolved  into  3  equal  factors  and  2m  of  them    taken.      Therefore 
(Sa**)"*  =  3™a  3 . 

3.  Affect  16a'5^~V with  the  exponent  f;  that  is,  represent 
2  of  the  3  equal  factors  of  16a^^x~^. 

Explanation. — I  Avill  first  resolve  IGa^^x-^  into  3  equal  factors  (or 
indicate  it  when  I  cannot  perform  it).  To  do  this  I  take  one  of  the  3 
equal  factors  of  16,  which  I  represent,  as  I  cannot  resolve  it,  and  write 

L 

(16)'*.    Again,  one  of  the  3  equal  factors  of  a^  *  is  a^,  as  a^  ^  is  15  factors 
of  a,  and  consequently  when  resolved  into  3  equal  factors  one  of  the  3 

contains  5  factors  of  a.     Thus  a  X  «  X  « to  15  factors  when  put 

into  3  equal  groups  becomes  aaaaa  X  aaaaa  X  aaaaa,  one  of  which  is 

a'.      To  resolve  «— ^  I  consider  it  as  -  =-  X-X—    X-X  —  X  -• 

♦C  »C  »l/         tC-  »c  *c  »c 

Separating  this  into  3  equal  groups  it  becomes  —^X—^X-^',    hence 

11 

one  of  the  3  equal  factors  of— -r  is-,  or  x~^.      Therefore  16a '"iir-^ 
^  x*'      x^ 

L 

being  resolved  into  3  equal  factors,  one  of  them  is  {lG)'^a^x-^.     But  I 
am  to  take  2  of  these,  as  the  exponent  f  indicates.     This  gives  me  (re- 

peat  the  reasoning  of  Ex.  1)    (16)-^a^"c— *. 

4.  Affect  3a"6~^  with  the  exponent  7 ;  that  is,  take  m  of  the 

n  equal  factors  of  it. 


INVOLUTION.  147 


Explanation.  3  n  represents  one  of  the  n  equal  factors  of  3.  One 
of  the  n  equal  factors  of  «»  is  a,  as  a"  means  n  factors  of  a.     6-"'  =: 

-=-XrX7-X torn  factors.     Each  of  these  being  separated 

h'^      b      b      b 

into  n  equal  factors  7X-X- to  n  factors  =  —.      And  taking 

b~   b~^    b^ 

one    of   the    factors   - ,    from    each  of   the  m  factors    -,  I  have  m 
J.-  ^ 

On 

factors  of  —  or  —    r=  5  «  .      Therefore  one  of   the  n  equal  factors  of 

]_  m 

3^»5— m  is  ^~ab    "  •     But  I  am  to  take  m  such  factors  which  gives  (re- 

m  m" 

peat  the  process  of  Ex.  2.)  3"a"'b    "  • 

[Note. — These  specimens  of  analysis  are  given  to  show  how  each 
example  should  be  shown  to  depend  for  its  solution  upon  the  elemen- 
tary principles.  Let  it  be  made  sure  that  the  pupil  can  do  this  in  any 
given  case,  after  which  he  should  simply  apply  the  rule,  until  he  be- 
comes expert  in  doing  it.  ] 

5.  What  is  the  square  of  -aPm  ?     Of  —  - — ,or  —  -  abx  ~'  ? 

2a'-^x  2  J^    _i 

The  cube  of —,  or —  -a'^m  ^x? 

3m^  ^ 

2 

6.  Affect  8ai2^~^  with  the  exponent  -.  jt    a  s    -* 

4 

7.  Affect  32a2o^i5  ^rith  the  exponent  —  -. 

Besidty 


8.  Affect  13x~'^y~^  with  the  exponent  —  3. 


148  POWERS   AND   ROOTS. 

9.  Perform  the  following  operations  and  explain  each 
as  a  process  of  factoring :  {d2a^^b-'')~^ ,  (lOOfl-^^*)"^, 
(IW^x'^y-'f^,  and  (a"'6rj;-^)~T. 

Results,  — — ,  10 — ,  ^— ,  and. 


^^'        ^     (11)%  aTh-^ 


19S,  J^rob.  3.  To  eximnd  a  binomial  affected  ivith 
any  exponent. 

It  ULE. — This  rule  is  best  stated  in  a  formula.  Thus,  let 
a,  6,  and  m  be  any  numbers  whatever,  positive  or  negative, 
integral  or  fractional,  then  will  (a  +  6)"*  represent  any 

BINOMIAL,  AFFECTED  WITH  ANY  EXPONENT,  AND 

1  -  /& 

mini  —  l)(m — 2)       ,,„ 
+       1-2-3     ^'     "'^ 

m{m-l)  (m-2)  (m-3) 
+  1-2-3-4  ""     ^ 

m{m-l)  (m-2)  (m-3)  (m-4) 

+  1-2-3-4-5  ''     ^  "^  ®^^- 

Dem. — This  formula  is  the  celebrated  Binomial  Theorem  discovered 
by  Sir  Isaac  Newton,  who  also  demonstrated  it.  There  are  several 
elementary  demonstrations,  but  they  are  somewhat  prolix  and  difficult 
for  young  students,  and  the  author's  experience  has  satisfied  him  that 
little  or  no  good  comes  to  most  students  from  their  study.  The  dem- 
onstration is,  therefore,  not  inserted  here,  but  will  be  found  in 
Appendix  I.  If  the  pupil  learns  the  formula,  and  learns  to  apply 
it  with  facility,  it  is  all  that  is  thought  best  for  him  to  attempt  at 
this  stage  of  his  progress. 

EXAMPLES. 

1.  Expand  {x-\-yY  by  the  Binomial  Formula. 

Model  Solution.— To  apply  the  B.  F.,  a;  =  a  of  the  formula, 

5  /^ ^\ 

y  =  h,   and  5  =  w.      .'.     I  have  {x  +  yY  =  a;*  +  53^~^y  +         — 


INVOLUTION.  149 

6    2  2.    5(5  -  1)(5  -  2)    ^     3  ^  ,  5(5-l)(5-2)(5-3)    ,     .   ^ 

5<5  — 1)(5  — 2)(5— 3)(5  — 4)     .    ,   -     ,      ,.^  ,         .i      ,       i 
-j — : x'^—^y",  at  which  term  the  develop- 

ment  becomes  complete  since  the  next  coefficient  would  hare  a  factor 
5  —  5  =  0  which  would  destroy  the  term.  Perforcaing  the  operations 
indicated  I  have  (x  +  t/)^  =  x^  -f  5x*y  -\-  lOx^y^  -f-  'i-Ox^y*  +5.r?/-* 
-\-  y^.  (In  practice,  this  result  should  be  written  out  without  writing 
the  preceding,  by  simply  applying  the  formula  mentally. ) 

2.  Expand  (x  —  ijy  by  the  B.  F. 

SuGOESTioNs.  ic  =  a,  —  2/  =  ^  and  G  =  m.     .  •.     (x  —  y)^  =  x^  -\~ 

6.=(  -  ,)  +  1^..(  _   ,).  +  ^    x^(  -   ,)»    +  ^.^ 

,^    .   6-5-4-3.2    ^         ^,    ,    6.5-4-3-2-1      „^ 
^'(-y)^+    2.3.4.5    ^(-y)^+       2.3-4.5.6-    -'(  -  V^'   = 
a;6  _  6x^2/  +  15x V  —  20x^2/'  +  ISx^y^  —  6xi/=  +  y^. 

3.  Expand  {2m"-  —  3n^)^  by  the  B.  F. 

Suggestions. — Make  a  =  2m*,  6  = —  Sn""'  and  m  =  i^      .«.      We 
have    (2?7i*  —  3n^)*  =  (27n2)4  -f-  4  (2m2)3(  —  3n')  -|-  ^-^   (2m2)2 

^-^'^'^'  +  i^(2m*)(-3n^)^  +  ^2^-^'^'^''^°^  -  ^^'^'    = 

1 
(by  performing  the  operations  indicated)  16m*  —  96m^n*  -f-216m'*?i  — 

216j;i2/i*  +  Sin*. 

4  Expand  {x  -{- y)-*  hj  the  B.  F. 

Suggestions,     a  =  x,  b  =z  y,  and  m  =  —  4.     .  • .     (x  -f-  y)~^  =  x  — ' 
,    .      ..   _i    1      ,    -4(— 4-1)        ,    ,  ,    ,  _4(-4-l)(-4  — 2^ 

x—^—'y^  -j-  etc.  This  series  does  not  terminate  since  no  factor  ever 
becomes  0  ;  but  the  development  can  be  carried  to  any  desired  extent. 
Performing  the  operations  indicated,  we  have  (x  -j-  y)~~^  =  x  — *  —4x--'y 
+  10x-«2/2  _  20x--!y^  +  35x-s?/*  — ,  etc. 

2. 

5.  Expand  (m  —  ?i)^  by  the  B.  F. 
Suggestions. — Making  the  proper  substitutions  in  the  formula,  we 


3.  2.  S.. 

■3  -  ' 


^(|-l)„f-^ 


have   (m  —  n)'^  =  7?i    -f- fm        (— n)   -j-  ^^^^ — m       (  —  n)^    + 


150  POWERS   AND   ROOTS. 

llt^Mri™J-V,.)»  +  MfMz!Hi^J-(_„).  +, etc.  A 

series  which  ruever  terminates,  since  no  coefficient  reduces  to  0.     Per- 

2.  % 

forming  the  operations  indicated  we  find  that  (m  —  nY  =   m*  — 

—  4-  —  i  -7-  - 1^ 

|m    ^n  —  ^m    ^n^  — A  ^     ^^^  —  F4U^        '^'^  — >  ^^c. 

6.  Expand  (1  +  ^)"  by  the  B.  F. 

Bes.,  (1  +  07)"=  1  +  wa;  +  -^—^ — ^-x^  +  '2~ 8      ^' 

n(n-l)(n-2)(n-3)  _  _   n(n-l) 

+  2^7-3—7—4         ^  +'  ^*^-  2        "^ 

ScH. — This  expansion  is  in  itself  a  very  useful  formula,   and  should 
be  memorized. 

7.  Expand  (3  —  y')^  by  the  B.  F. 

-J.  _3  -5 

JSeswZ^,  (3  —  2/2)2  3^  32  _  f —      _  £_  y.  _  £ e  _ 

5-3"' 


128  -y'—^^^' 

8.  Expand  (1  +  x^-y  by  the  B.  F. 

Besult,  1  +  5^2  +  10.27^  +  10a;c  +  5x»  +  a;io. 

9.  Expand  (1  —  a'^)~^  by  the  B.  F. 

■10  pr  OPT 

Besult,  1  +  -  a2  +  _  a.  4-  —  a«+  —  a8+,  etc. 


10.  Expand  v^a^  —  as^a  by  the  B.  F. 

•i     Besult,  ^a-  —    a'^e'^  =  a^l  —    e^    =  a{X   —  62)"2  = 
^  ,,        1  1  1-3  13-5 

<^~r'-Y:i''- 27176 '' ~  2:7:^ ^'  -^  ^^-)- 

11.  Expand  - — ^——  by  the  B.  F. 

Besult, =  (c  +  xy  =:  0-^—20 ~*x  +  3c -*x^  — 

{c-^  xy      ^        ^ 


INVOLUTION.  151 

4:C-'x^  +  &c.  =  —(1 + +,  etc.). 

12.  Expand 1  by  the  B.  F. 


{1  +  x) 


Besult, 1    =    (1  +  ^)        =1 -\ 

(1  +  X?  5         25 

ll;r3        44a:4 

1 ,  etc. 

125    ^625 

13.  Expand  (a;  +  2/  +  c)^  by  the  B.  F. 

Suggestions.— Put  (a;  +  y)  —  z  .-.  (x  4-  2/  +  c)*  =  (z  +  c)*  = 
z*  -^  423c  -j-  63* c'*  4-  "^c^  -f-  c*  =  (restoring  the  value  of  z) 
(X  +  y)*  +  4(x  +  yYc  +  6(x  +  y)2c^  +  4(x  +  !/)c'  +  c^  But 
(X  +  y)*  =  X*  -f  4xV  +  6x*y*  +  4xy^'  +  y\  (x  +  v)'  =  x=^  +  3x«y 
-j-  3xy*  4-  y-',  and  (x  +  y)^  =  x*  +  '^*!/  +  y^-  Whence  by  substitu- 
tion we  have  (x  -f-  2/  4-  c)*  =  x*  +  4x^y  +  Gx^y*  -\-  4xy^  +  y*  + 
4cx=»  +  12cx*y  4-  12cxy2  -f-  4cy^  +  6c«x«  -f  12c'xy  4-  Gc^y^  -f  4c^x 
-}-  4c^y  +  c*. 

14.  Expand  (2a  —  6  +  0^)3  by  the  B.  F. 

Besuli,  8a3  —  12^26  4-  606^  _  fcs  4.  12^202  —  12a&c2  + 
362C2  4-  6ac^  —  36c^  4-  c^. 

100.  Cor.  1. — T/2e  expansion  of  a  hinomial  teiminates 
only  when  the  exponent  is  a  positive  integer,  since  only  when 
m  is'  a  positive  integer  will  a  factor  of  the  form  m(m  —  1) 
(m  —  2)(m  —  ?>)etc.  become  0,  as  is  evident  by  inspection. 

19  7  >  Cor.  2. — When  m  is  a  positive  integer,  that  is  when 
a  binomial  is  raised  to  any  power,  there  is  one  more  term  in 
the  development  than  units  in  the  exponent.     Since  the  first 

coefficient  is  1;  the  2nd,?7i;  the  3rd,  — — --,  the  4th, 

m{m  —  l)(m —  2)     .         ,     m{m  —  l)(m  —  2)(m —  3) 
__         _        ;  the  5th, 2        •       3 ^-l        ' 

i&c,  we  notice  that  the  last  factor  is  m  —  (the  number  of 
the  term  —  2) ;  and  the  number  of  the  term,  therefore,  which 


152  POWERS   AND   ROOTS. 

lias  m  —  TO  as  a  factor  is  the  (m  +  2)th  term.  But  this 
is  0.     Hence  the  (m  +  l)th  term  is  the  last. 

108.  Cor.  3. —  When  m  is  a  positive  integer  the  coefficients 
equally  distant  from  the  extremes  are  equal;  since  (a  +  6)"*  = 
(5   _|_   a)"»  ;    the   former   of    which    gives   a"*  +  moT-^b  + 

'^^i''''—^)  ar-'^y^ 4.,  etc.,   and  the  latter  6"*  +  mb'^-'a  -\- 

— 1^ — I  5'»-V  _^j  qIq^     "Whence  it  appears  that  the  first 

half  of  the  terms  and  the  last  half  are  exactly  symmet- 
rical. 

100.  Cor.  4. — The  sum  of  the  exponents  in  each  term  is  the 
same  as  the  exponent  of  the  power. 

ScH. — The  last  two  corollaries  apply  to  the  form  (x  -\-  y)"",  and  not 
to  Buch  forms  as  (2a"'  —  36'^)'",  after  the  latter  is  fully  expanded. 

200.  Cor.  5. — A  convenient  rule  for  writing  out  the  pow- 
ers of  binomials  may  he  thus  stated : 

1st.  Tliere  is  one  more  term  in  the  development  than  there 
are  units  in  the  exponent  of  the  power. 

2nd.  The  first  contains  only  the  first  letter  of  the  binomial^ 
and  the  last  term  only  the  second,  while  all  the  other  terms  con- 
tain both  the  letters. 

Srd.  The  exponent  of  the  first  letter  of  the  binomial  in  the 
first  term  of  the  development  is  the  same  as  the  exi^onent  of  the 
required  power  and  diminishes  by  unity  to  the  right,  tvhile  the 
exponent  of  the  second  letter  begins  at  unity  in  the  second  term 
of  the  expansion  and  increases  by  unity  to  the  right,  becoming, 
in  the  last  term,  the  same  as  the  exponent  of  the  jMiver. 

Uh.  The  coefficient  of  the  first  term  of  the  expansion  is 
unity  ;  of  the  second,  the  exponent  of  the  required  power;  and 
that  of  any  other  term  may  he  found  by  multiplying  the  coeffi- 
cient of  the  jjreceding  term  by  the  exponent  of  the  first  letter  in 
that  term,  and  dividing  the  product  by  the  exponent  of  the  sec- 
ond letter  +  1. 


INVOLUTION.  153 

Dem.— This  rule  is  a  deduction  from  the  formula   (a  +  6)™  = 

m{m  —  l)(m  -  2)(m  -  3)  ^^_^^  _^ ^^ 

2-3-4:  ^ 

1st.  Is  proved  in  Cob.  2,     Eepeat  it. 

2nd.  Is  proved  in  Cob.  3.  Kepeat  it. 

3d.  The  law  of  the  exponents  is  directly  observable  from  the 
formula. 

4th.  The  coefficients  of  the  first  and  second  terms  are  seen  in  the 
formula  to  be  as  stated.  The  coefficient  of  the  third  term  may  be  writ- 
ten m  X  — : .  which  is  the  coefficient  of  the  second,  or  preceding 

2 

term  (m),  multiplied  by  the  exponent  (m  —  1)  of  the  first  letter  in  that 

term,  and  divided  by  2  which  is  the  exponent  (1)  of  the  second  letter 

-f-  1.    In  like  manner  noticing  any  other  term,  as  the  5th.    Its  coefficient 

m(m  —  l)(m  —  2)       ?n  —  3               m{m  —  l)(m  —  2) 
may  be  written — -  X  — j •     l^ut ^ — 

is  the  coefficient  of  the  Ith  term,  m  —  3  is  the  exponent  of  the  first 
letter  in  the  4th  term,  and  4,  the  divisor,  is  the  exponent  of  the  sec- 
ond letter  (3)  -f  1. 

15  to  20.  Write  out  by  the  above  rule  the  expansions  of 
the  following:  {m  +  n)',  {x  +  y)\  {a  +  cy,  {x  +  my, 
2  X 

%  1 

Suggestions  upon  the  last.— Eegard  a'  and  m^  as  simple  num- 

2.  JL 

bers  represented  by  letters  without  exponents.     Thus  (a'*  -\-  vi^)^  = 

(a^)4  _|.  4  {(^^{m^)  4-  6  (a')2(7n*)2  4-4  (a^)(m'^)»+  (m^)*.  Nowper- 

^1  ^  fi    -1- 

forming  the  operations  indicated,  we  have  (a"*  ^-^'^  )*  =  «^  +  4a"^r/i^ 

i.  i    3. 

-f  6aSji  +  4aS7i^  +  m^. 

201,  CoR.  6. — If  the  sign  between  the  terms  of  the  bino- 
mial is  minus,  «s  (a  —  b)'",  the  odd  terms  of  the  expansion 
are  -\-  and  the  even  ones  — .  This  arises  from  the  fact  that  the 
odd  terms  involve  even  powers  of  the  second  or  negative  term 
of  the  binomial,  and  the  even  terms  involve  the  odd  powers  of 


154  POWERS   AND   ROOTS. 

the  same.  Thus  the  second  term  involves  ( —  h)  which 
makes  its  sign  —  ;  the  4th  term  has  ( —  hy,  the  6th  term 
( —  hy,  &c.  But  the  first  term  does  not  involve  ( —  6),  and 
the   3rd   has   ( —  hy  or   h\  the  5th  has  ( —  6)^  or  b*,  etc. 

21  to  24.  Write  out  the  expansions  of  (m  —  n)^,{x  —  y)', 
(^2 —  ^3)3^  (a^—  h'^y. 

Besult  of    the   last,    (a^—  6^)^  =  a^  — 4a%^4- 6a6^— 

4:ah  +  Z>^. 


SECTION  II. 
Evolution. 

202,  I^TOh,  1,  To  extract  any  root  of  a  perfect  power 
of  that  degree. 

R  ULE. — Resolve  the  number  into  its  prime  factors,  and 

SEPARATE  these  INTO  AS  MANY  EQUAL  GROUPS  AS  THERE  ARE  UNITS 
IN  THE  DEGREE  OF  THE  ROOT  REQUIRED  ;  THE  PRODUCT  OF  ONE  OF 
THESE  GROUPS  IS  THE  ROOT  SOUGHT. 

Dem. — Since  the  mth  root  {i.  e.,  any  root)  of  a  number  is  one  of  the 
m  equal  factors  of  that  number,  if  a  number  is  resolved  into  m  equal 
factors,  as  the  rule  directs,  one  of  them  is  the  mth  root. 

EXAMPLES. 

1.  Extract  the  cube  root  of  74088. 

Model  Solution. — Resolving  74088  into  its  prime  factors  I  find 
them  to  be  2 •2.2.3-3 -3. 7- 7- 7.  These  arranged  in  3  equal  groups  give 
2.3.7  X  2.3.7  X  2.3.7.  Hence  2.3.7  =  42  is  the  cube  root  of  74088, 
since  it  is  one  of  its  3  equal  factors. 

2  to  5.  Extract  in  this  manner  the  following  :    v/492804, 

^592704,  V248882,  v' 456583.  ^^ots,  702,  84  and  12. 

2  _i 
6.  Extract  the  square  root  of  Ula''x~^y'^z    s. 

Model  Solution. — The  two  equal  factors  of  81  are  9  •  9  ;  of  a*, 
a*  'Ci^  ;    of  05-2,  x-i  -x-i  ;  of  ^/^  y*  -  y^ ;  of  z~^  ,  z~  ^  '^  .  z~     .    Hence 


EVOLUTION.  155 

81a*xr-Yz      =  9a^x~yz~^X  Qa^x-^z'^,  and      consequently, 
da'ar-^z    ^°  is  its  square  root. 

7  to  11.   Extract  ^^25ai^,  J64a-'A  ^^^xy^,    ^14Aa<)n', 

l^a^  3.  Ill 

SJsGrrm     ^^^^^''   ±  ^^'^''  ±8a-'x^,  ±lx^y^,  +12aW, 


I 


12  to  15.    Extract   Vl25m6a;»2,    3  Ul^^x'^y^,     ^—  'd'la'^y -\ 

2    JL  _  1 

JSoo^s,  ^m^x\  12j;^2/^>  —  Sa^;?/-^,  and  4-  2n  "^^/^ 

QuEEY. — Why  the  ambiguous  sign  to  the  last? 

203,  ScH. — The  sign  of  an  even  root  of  a  positive  number  is  am- 
biguous (that  is,  +  or  — )  since  an  even  number  of  factors  gives  the 
same  product  whether  they  are  positive  or  negative  {87 f  88).  The 
sign  of  an  odd  root  is  the  same  as  that  of  the  number  itself,  since  an 
odd  number  of  positive  factors  gives  a  positive  product  and  an  odd 
number  of  negative  factors  gives  a  negative  product  {88,  89). 

204»  Cor.  I. — The  roots  of  monomials  can  be  extracted  by 
extracting  the  required  root  of  the  coefficient  and  dividiyig  the 
exponent  of  each  letter  by  the  index  of  the  root,  since  to  extract 
the  square  root  is  to  affect  a  number  with  the  exponent  ^,  the 
cube  root  -^,  the  nth  root  7,  etc.     (194,) 

EXAMPLES, 

16  to  21.  In  this  manner  write         J25a'^b",  ^ —  343;zr^?/~*, 

Nj  ^'  \Q4.m^y<^  N     243%-'" 

f 

Roots,   ±  5a\  —  Ix'^y-'   ±  Sm'-n'^x^,  3c^m~^V^,   ^-^ 

^  '   4?7iy^ 

2a^x^ 
and  — • 

'dbiy-' 


156  POWERS   AND   ROOTS. 

205,  Cor.  2. — The  root  of  the  product  of  aexjeral  numbers  is 
the  same  as  the  product  of  the  roots.  Thus,  '^yabcx  = 
v/a  -  Vft  •  v/c  •  'Vx,  since  to  extract  the  mth  root  of  abcx  we 
have  but  to  divide  the  exponent  of  each  letter  by  m,  whicii 

1    1  2.  1 

gives  fl»«5m^m^r«  ^^  ^^  .  V^  '  Vc  -  Vx. 

206.  Cor.  3. — The  root  of  the  quotient  of  two  numbers  is  the 

same  as  the  quotient  of  the  roots.     Thus,  ^  1^  is  the  same  as 

'Sjn 

-~,  since  to  extract  the  rth  root  of  H}.  we  have  but  to  ex- 

tract  the  rth  root  of  numerator  and  denominator,  which  oper- 
ation is  performed  by  dividing  their  exponents  by  r.     Hence 


\m        m~ 


1  _ 

vm 


V\ 


EXAMPLES. 


22.  Show  that  ^8  x  27  =  v/8  x  ^21. 

MoDEjj  Solution.— We  may  show  this  in  two  ways.  1st, — 
Experimentally.  Thus  1/8  X  27  =  ^"216  =  6.  Again  ?/8~  X 
^27"=  2X3  =  6.  Hence  ^8  X  27  (or  the  cube  root  of  the  product)  = 
\/8  X  v^27  (or  the  product  of  the  cube  roots).  2nd, — Analytically. 
\/8  X  27  signifies  that  the  product  of  8  and  27  is  to  be  resolved  into 
3  equal  factors,  which  is  accomphshed  by  resolving  8  into  its  prime 
factors,  and  27  into  its  prime  factors,  and  then  separating  these  factors 
into  three  equal  groups  (202).  This  will  .give  the  same  result  as  resolv- 
ing the  product  of  8  and  27,  or  216,  into  3  equal  factors,  since  the  prime 
factors  of  216  are  the  same  as  those  of  8  and  27. 


Ja-"'67  =  ^«—  Xllbn 


23.  Show  that  ..  la-"'6« 

Suggestions. — The   5   equal  factors  of  a-"'6''  are  a   ^  &^",  a   ^  P^^'j 
a   Fb^n^  a~r&s^,  and  a    Th^^,  since  by  the  rules  of  multiplication  these 


EVOLUTION.  167 

1  m    1 

multiplied  together  make  a-^h~.    But  a"  '^b^  is  the  product  of  one  of  the 
5  equal  factors  of  a-"*,  or  Va-"\  and  one  of  the  5  equal  factors  of 

-         fT 

6",  or    °  I  5«  • 

24.  Extract  the  square  root  of  a^C^  -\-  2a^bc^  +  a-b'^cK 

Solution. — The  factors  of  this  are  readily  seen  to  be  a^,  c^,  and 
4r+2a6  +  6^  which  separated  into  two  equal  groups  give  ac{a+b)  and 
ac{a+b).     Hence  ac{a  +  b)  or  a^c  +  abc  is  the  required  root. 

25.  Extract  the  square  root  of  m^  —  2m^x  +  'in^^^. 

Booty  ?7i*(l — •  a:)  or  m-  —  m^x. 
ScH. — The  extraction  of  roots  by  resolving  numbers  into  their 
factors  according  to  this  rule,  is  limited  in  its  apphcation  for  several 
reasons.  In  the  case  of  decimal  numbers  we  can  always  find  the  prime 
factors  by  trial,  and  hence  if  the  number  is  an  exact  power,  can  get  its 
root.  But  in  case  the  number  is  not  an  exact  power  of  the  degree  re- 
quired, we  have  no  method  of  approximating  to  its  exact  root  by  this 
rule,  as  we  have  by  the  common  method  already  learned  in  arithmetic. 
In  case  of  Hteral  numbers  the  difficulty  of  detecting  the  poljmomial 
factors  of  a  i^oljTiomial  is  usually  insuperable.  Hence  we  seek  general 
rules  which  ynR  not  be  subject  to  these  objections. 


207*   I* rob,  2,    To  extract   the  f<quare  root  of  a  poly- 
nomial. 
RULE. — 1st.  Aeeange  the  polyng^mial  with  eefeeence  to 

ONE  OF  ITS  LETTERS,   AS  FOR  DIVISION. 

2nd.  Extract  the  square  root  of  the  first  left  hand 
TERM.  This  root  is  the  first  term  of  the  required  root. 
Subtract  the  square  of  this  term  of  tke  root  from  the  pol- 
YNOivn  A  r.. 

3rd.  Double  the  root  already  found  for  a  Trial  Divisor. 
By  this  trlal  divisor  divide  the  first  term  of  the  remainder 
of  the  polynomial,  and  "write  the  quotient  as  the  second 
term  of  the  root. 

4:th.  Complete  the  divisor  by  adding  to  the  trlal  divisor 

THE  last  term  IN  THE  ROOT.       MULTIPLY  THE  TrUE  DiVISOR  THUS 
formed  by  THE  LAST   TERM  IN  THE  ROOT,  AND  SUBTRACT  THE  PRO- 


158  POWERS   AND   ROOTS. 

DUCT  FROM  THE  LAST  BEMATNDER,  BRINGING  DOWN  SUCH  TEEMS  AS 
MAY  BE  NECESSABT. 

5tli.  Double  the  boot  aleeady  found,  as  a  new  Tbial  Di- 
visoe  ;  divide  the  last  bemaindeb  by  it  ;  complete  it  by 
adding  to  it  the  last  tebm  ;  multiply  and  subtract  as  before. 
Continue  to  repeat  the  process  op  forming  Trial  Divisobs, 

DIVIDING,  completing  THE  DIVISOR,  MULTIPLYING  AND  SUBTRACT- 
ING, IN  THE  SAME  WAY  TILL  THE  POLYNOMIAL  IS  EXHAUSTED,  OB 
UNTIL  THERE  IS  NO  TEEM  OF  IT  BEMAINING  WHICH  CAN  BE  EXACTLY 
DIVIDED  BY  THE   FIRST  TERM  OF  THE  TRIAL  DIVISOR. 

Dem. — 1st.  The  poljraomial  is  arranged  as  in  division,  since  such, 
is  the  order  which  the  terms  assume  in  squaring  any  polynomial  root. 

2nd.  In  squaring  any  polynomial,  the  first  term  of  the  square  is 
found  to  be  the  square  of  the  first  term  in  the  root ;  hence,  in  extracting 
the  square  root,  the  square  root  of  the  first  term  in  the  given  poly- 
nomial is  the  first  term  in  the  root. 

3rd.  To  prove  the  process  of  finding  the  divisors  and  the  subsequent 
terms  of  the  root,  we  observe  the  following  operations  : 

A.  (a  +  6)2  r=  a2  4.  2a6  -\-h^  =  a^  +  (2a  +  h)b. 

B.  (a  4-  6  -f  c)2  =  [(a  +  6)  +  cV  =  («  +  ^V  +  i^i^^  +  &)  +  c]c 

=  a 2  4-  (2a  +  6)6  4-  [2(a  -f  6)  -f  c]c.* 

C.  (a  +  6  -f  c  4-  d)2  ==  [(a  +  6  -f  c)  +  d]2  =  (a  +  6  4-  c)«  4- 

[2^a  4-  6  -f  c)  +  did  =  a"  4-  (2a  4-  6)  6  + 

[2(a  4-6)  +  c]c  4-  [3(a  -f  6  4-  c)  4-  d]d. 
Hence  it  appears  ;  1st,  That  the  square  of  a  polynomial  (as  a  -{-  h, 
a  4-  &  +  c,  or  a  -{- h  -\-  c  -}-  d)  is  made  up  of  as  many  parts  as  there 
are  terms  in  the  root  ;  2nd,  That  the  first  part  is  the  square  of  the 
first  term  in  the  root ;  3rd,  That  the  second  part  is  Twice  the  first 
term  of  the  root  (the  part  already  found),  4"  the  second  term  of  the 
root,  multiplied  by  the  second  term  ;  4th,  That  any  one  of  these 
parts,  as  the  nth,  is  Twice  that  portion  of  the  root  previously  found,  or 
Twice  the  n  —  1  preceding  terms  of  the  root,  4-  the  nth  term  of  the 
root,  multiplied  by  this  last  or  ?ith  term. 

*  This  expansion  is  made  by  squaring  [{a  -{-  b)  -{-  c],  as  {a  -\-  b)  is  in  the  result 
above.  That  is,  for  a  substitutiug  a  -\- b,  and  for  b,  c.  This  gives  [{a  +  6)  +  c]2  = 
(o  _^  5)2  -j-  [2(a  -|-  b)  +  c]c.  Then  for  (a  +  6)2  substituting  its  value  as  found  iu 
A,  we  have  (a  +  b  +  c)2  =  a"*  +  {2a  +  b)b  +  [2  (o  +  6)  +  c]c. 


EVOLUTION.  159 

ScH.  1. — K  the  first  term  of  the  arranged  polynomial  is  not  a  perfect 
square,  the  root  can  not  be  extracted. 

ScH.  2.— If  at  any  time  no  term  of  the  remainder  can  be  exactly 
divided  by  the  first  term  of  the  trial  divisor,  the  root  can  not  be 
extracted. 

EXAMPLES. 

1.   Extract  the  square  root  of  49^2^^  —  SOax^  +  25x-*  -f 
16a*  —  24^3  J7. 

MODEL   SOLUTION. 
OPEEATION. 

16a*  —  24a3x  +  49a««*  —  SOax^  +  25x*   I  4a«  —Sax-^5x^. 

16a;; ' 

8a'  —  dax     I  _  24a^x -f- 49a--^x-^ 
—  24a''.r-f-    9a  2^2 
8a-^  —  6aa;  -f  5.p^    |  -iOa^x''  —  SOax-^  +  25a;* 
■ '  40a^.x-^  —  30a.r;^^  +  25x* 


Explanation. — If  this  polynomial  is  a  perfect  square,  the  term 
containing  the  highest  power  of  a  or  x,  is  the  square  of  the  first 
term  in  the  root.  Hence  I  place  lea-*  first  in  the  arrangement.  (25a;* 
would  do  as  well.)  And,  as  the  terms  arising  from  squaring 
a  polynomial  (as  the  root  of  this  given  number),  are  arranged  ac- 
cording to  the  leading  letter  of  the  root,  I  arrange  the  given  polyno- 
mial according  to  the  powers  of  a,  as  this  wiU  be  the  leading  letter  of 
the  root,  when  I  put  16a^  first  in  the  power. 

Having  arranged  the  given  number,  I  know  that  the  square  root  of 
the  first  term,  16a^,  or  4a2  is  the  first  term  of  the  root,  since  in  squaring 
any  polynomial  (as  the  root  sought),  the  first  term  in  the  square  is  the 
square  of  the  first  term  in  the  root. 

Now,  having  removed  by  subtraction  the  square  of  the  first  term  of 
the  root,  I  double  the  root  already  found,  obtaining  8a^  for  a  trial 
divisor.  This  is  the  trial  divisor,  since  the  second  part  of  any  square 
(the  square  of  the  first  term  of  the  root  being  called  the  first  part)  is 
twice  the  root  already  found  -\-  the  next  term  of  the  root,  multiphed 
by  this  next  term,  I  now  find  that  the  trial  divisor  goes  into  the  first 
term  of  the  remainder  —  3aa;  times,  which  is,  therefore,  the  second 
term  of  the  root.  But  the  true  divisor  is  twice  the  root  previously 
found  plus  this  last  term  ;  hence  I  add  —  3aa;  to  the  trial  divisor  and 
have  8a2  —  Sax  as  the  true  divisor.  Multiplying  this  true  divisor  by 
the  last  term  of  the  root  I  obtain  —  24a"a;  -|-  da-x-,  which  is  the  second 


160  POWEKS    AND   ROOTS. 

part  of  the  given  power.  [lu  the  demonstration  it  is  shown  by  squaring 
a  polynomial,  that  the  second  part  of  the  square  is  twice  the  root  pre- 
^iously  found  -f-  the  next  term  of  the  root  (the  true  divisor) 
multiphed  by  this  next  term,  or  the  one  just  found.  That  is,  in  this 
case  (8a2  —  3ax)  3ax.  ] 

Subtracting  this  second  part  of  the  power,  the  third  part  is  twice 
the  root  already /owncZ.  -f-  the  next  term  of  the  root,  multiphed  by  this 
next  term.  In  this  case,  therefore,  the  new  trial  divisor  is  Sa^  — Qax. 
[Proceed  just  as  in  the  last  paragraph,  and  complete  the  expla- 
nation. ] 

Finally  4a2 — 3aa; -|-  ^^^  i^  ^^^  required  root,  for ,'^  indeed,  I  have 
actually  squared  it  and  subtracted  this  square  from  the  given  num- 
ber and  found  no  remainder.  [The  pupil  should  notice  that  the 
sum  of  the  several  subtrahends  is  the  square  of  the  root.  ] 

[Note. — The  pupil  should  explain  in  full,  each  of  the  following 
examples,  giving  the  reason,  as  above,  for  every  step.  In  this 
manner  it  is  hoped  that  the  method  will  be  seen  as  an  argument, 
instead  of  merely  a  process  of  operating.] 

2.  Extract  the  square  root  of  8^  +  4  +  :r^  -f-  4^'  +  8^1 

Root,  x'  +  2x  +  2. 

3.  Extract  the  square  root  of  25a^^'  —  12a^''  +  16a^  -f 
4^*  —  24a'^.  Root,  Ix"  —  ^ax  +  4a^ 

4.  Extract   the    square   root   of  x^  —  ^ax^  -f   15a^x*  — ■ 

X'i 

5.  Extract  the  square  root  of  ^-^  —  x^  -\-  -  -f  4^  —  2  -|- 

^  «    .  ^       2 

-  .  Root,  x^  —  -  H — 

x^  '  2       X 

*  -L    2  X 

G.  Extract   the    square   root   of  9x  —  24.r'^7/^  +  12^^  -|- 

IQy^  —  16^^  +  4.  Root,  Sx^  —  Itf  +  2. 

7.  Extract  the  square  root  of  9a-'  +  12a  "^6-  —  6a  +  46^ 

—  4a262  +  a\  Root,  3a "'  -f-  2b^  —  a\ 

Query.— In  arranging  the  last  with  respect  to  a,  why  should  45* 
come  before  —  G«  ? 


EVOLUTION.  161 

208,  ScH. — Since  the  square  root  of  a  quantity  is  either  -j-  oi 
^-,  all  the  signs  in  the  above  roots  may  be  changed,  and  they  -^ill 
still  be  the  roots  of  the  same  polynomials.  Thus,  in  the  3rd  Ex- 
ample, if  we  call  the  square  root  of  4x*,  minus  2x'^,  ( —  2x-,)  which 
it  is,  as  well  as  -j-  2x-,  and  then  continue  the  work  as  before, 
we  get  for  the  root  • —  2x2  _^  3^^;  —  4.a'. 


209,  J*rob,  3m  To  extract  the  Square  Root  of  a  Decimal 
Number  either  exactly  or  approximately. 

RULE. — 1st.  Skparate  the  number  into  periods  by  plac- 
ing A  MARK  OVER  UNITS  AND  OVER  EACH  ALTERNATE  FIGURE 
THEREFROM,     CAJiTiING     THE     MARKED      FIGURE     WITH     THE      ONE, 

AT  ITS  LEFT,  IF  ANY,  A  PERIOD.  The  number  of  figures  in  the 
root  is  equal  to  the  number  of  periods  thus  formed. 

2nd.  Take  the  square   root  of  the   greatest   square  in 

THE  left  hand  PERIOD,  AND  WRITE  IT  AS  THE  HIGHEST  ORDER 
IN  THE  ROOT.  SUBTRACT  THE  SQUARE  OF  THIS  FIGURE  FROM 
THE  PERIOD  USED,  AND  TO  THE  REMAINDER  ANNEX  THE  NEXT 
PERIOD    FOR   A   NEW   DRTDEND. 

3rd.  Double  the  root  already  found  for  a  Trial  Divi- 
sor, BY  WHICH  DIVIDE  THE  NEW  DFVTDEND,  REJECTING  IN  THE 
TRIAL  THE  RIGHT  HAND  FIGURE  OF  THIS  DIVIDEND.  ThE  QUO- 
TIENT IS  THE  NEXT  FIGURE  IN  THE  ROOT  OR  A  GREATER  ONE. 
To  OBTAIN  THE  TrUE  DiVISOR  ANNEX  TO  THE  TrIAL  Df^TISOR 
THE  LAST  ROOT  FIGURE.  MULTIPLY  THIS  TrUE  DiVISOR  BY 
THE  LAST  ROOT  FIGURE,  SUBTRACT  THE  PRODUCT  FROM  THE 
LAST  NEW  DIVIDEND,  AND  TO  THE  REMAINDER  ANNEX  THE  NEXT 
PERIOD    OF   THE    GIVEN   NUMBER    FOR   ANOTHER   NEW   DIVIDEND. 

4:th.  Double  the  root  already  found  for  a  new  Trlal 
Divisor,  and   repeat  the   processes  as    given   in  the   3rd 

PARAGRAPH.  CONTINUE  TO  REPEAT  THE  OPERATION  TILL  ALL 
THE  PERIODS  ARE  BROUGHT  DOWN.  If  THE  NUMBER  IS  A  PER- 
FECT SQUARE,  THE  LAST  REMAINDER  IS  ZERO.  If  THIS  IS  NOT 
THE  CASE,  ANNEX  PERIODS  OP  TWO  O's  EACH,  AND  CONTINUE 
THE  PROCESS.  TILL  THE  REQUIRED  DEGREE  OF  ACCURACY  IS  AT- 
TAINED. All  the  root  figures  arising  from  decimal  frac- 
tional PERIODS   ARE  DECIMAL  FRACTIONS. 


162  POWEES   AND    liOOTS. 

ScH.  1. — In  pointing  off  decimal  fractions,  or  the  fractional  part  of 
mixed  numbers,  make  full  periods  of  two  figures  each,  annexing  a  0 
if  necessary. 

ScH.  2. — If  at  any  time  the  Trial  Divisor  is  not  contained  in  the 
dividend  to  be  used,  according  to  the  3rd  paragraph  of  the  rule, 
annex  a  0  to  the  root  and  also  to  the  Trial  Divisor,  and  then  bring 
down  the  next  period  and  divide. 

ScH,  3. — "When  the  work  does  not  terminate  with  the  last  period  of 
significant  figures  it  will  not  terminate  at  all,  and  the  given  number  is 
a  surd.  This  is  evident,  since  the  process  makes  the  unit's  figure  in 
each  subtrahend,  the  square  of  the  last  figure  in  the  root,  but  no 
figure  squared  gives  0  in  unit's  place.  The  process  can,  however,  be 
carried  to  any  given  degree  of  accuracy. 

Dem. — 1st.  That  this  method  of  pointing  gives  the  number  of 
places  in  the  root,  is  made  evident  by  squaring  a  few  numbers. 
Thus  the  square  of  1  is  1,  and  the  square  of  10  is  100,  hence  the 
squares  of  all  numbers  between  1  and  10  have  1  or  2  figures ;  that  is. 
Twice  as  many  Jigures  as  the  root,  or  one  less  than  twice  as  many.  Again, 
the  square  of  100  is  10000 ;  hence  the  square  of  all  numbers  between  10 
and  100  have  3  or  4  figures ;  that  is,  Twice  as  many  as  there  are  in  the  root, 
or  one  less.  In  like  manner  it  is  readily  seen  that  the  square  of  any 
number  consists  of  7\oice  as  many  Jigures  as  the  root,  or  one  less.  Hence 
the  method  of  pointing  indicates  the  number  of  figures  of  which  the 
root  consists. 

In  the  case  of  decimal  fractions,  since  the  number  of  decimals  in  a 
product  equals  the  number  in  both  the  factors,  there  are  always  twice 
as  many  decimals  in  the  square  as  in  the  root.  Hence  if  the  number 
of  decimal  places  in  the  given  number  is  odd,  they  are  to  be  made 
even  by  annexing  0. 

2nd.  That  the  greatest  square  in  the  left  hand  period  is  the 
square  of  the  highest  order  in  the  root,  appears  from  the  facta  that 
the  square  of  any  number  of  units  between  1  and  9  falls  in  the  first 
right  hand  period,  the  square  of  any  number  of  tens  between  1  and  9 
falls  in  the  second  period,  the  square  of  any  number  of  hundreds  be- 
tween 1  and  9  falls  in  the  third  period,  etc.  Moreover,  though  the 
left  hand  period  usually  contains  more  than  the  square  of  the  highest 
order  in  the  root,  it  can  not  contain  the  square  of  a  unit  more  of  that 
order,  since  all  the  figm-es  that  can  follow  this  highest  order  in  the  root 
can  not  make  another  unit  of  this  order.  Thus  the  square  of  3999, 
can  not  be  as  great  as  the  square  of   4000  ;    hjit  the  square  of   4000 


EYOLUTION.  163 

gives  IG  in  the  highest  period  of    the  power,   hence   the   square  of 
3999,  must  give  less  than  16  in  that  period. 

3rd.  To  prove  the  method  of  finding  and  using  the  Trial  Divisor^ 
suppose,  in  any  given  case,  the  pointing  shows  that  the  root  consists  of 
4  figures.  Represent  the  thousands  of  the  root  by  T,  the  hundreds  by 
h,  the  tens  by  t,  and  the  units  by  u.     The  number  itself  will  be 

^T+h-{-t-j-ii)'  =  r^  -f-  (2r+  h]h-i-  [2(r+  h)  +t]t  -\~ 

[2(  r  +  /i  +  0  +  w]  U'     {207,  3rd  paragraph  in  the  Dem.  ) 

"Whence  having  found  and  removed  the  scpiare  of  the  thousands,  T^, 
the  next  part  of  the  power  is  (2  T"  -f-  ^)^-  ^^^  ^^  the  lowest  order  of 
this  is  hundreds  multiplied  by  hundreds,  or  10,000's  vre  need  not  bring 
down  anything  below  10,000's,  or  the  next  period,  for  none  of  this 
part  is  contained  in  the  lower  periods.  Again,  for  trial,  considering 
this  part  as  2  T  X  ^,  the  product  contains  nothing  lower  than  hundred 
thousands  ;  hence  the  ten  thousands  figure  is  to  be  omitted  in  the  trial 
division.  But  the  true  divisor  is  2  T  -f  7i, ;  hence  the  root  figure  is 
annexed  to  the  trial  divisor,  or  really  added  to  it,  regarding  the  local 
values. 

4th.  It  is  evident  that  this  process  is  merely  repeated  as  we  increase 
the  number  of  figures  in  the  root ;  and  as  the  law  of  notation  is  the 
same  when  we  pass  the  decimal  point  into  a  fraction,  no  special  exem- 
plification is  needed  in  such  a  case. 

EXAMPLES. 

1.  Extract  the  square  root  of  7284601. 

MODEL   SOLUTION. 

oPEBATioN.      •  7284601(2699 


46)328 
276 

529")5246 
4761 


5389)48501 
48501 


Explanation. — As  the  highest  order  in  this  number  is  millions,  the 
highest  order  in  the  root  is  thousands,  for  the  square  of  thousands  is 
millions  or  ten  miUions,  and  the  square  of  anything  above  thousands 
is  more  than  ten  millions.  The  root,  therefore,  consists  of  thousands, 
hundreds,  tens,  and  units,  as  indicated  by  the  pointing. 


164  POWEES   AND   EOOTS. 

Since  the  square  of  the  thousands  figure  in  the  root  is  in  the 
number,  it  must  be  sought  in  the  7  milhons,  or  the  left  hand  period. 
The  greatest  square  in  this  being  4,  the  square  root  of  which  is  2,  the 
thousands  figure  of  the  root  is  2,  which  I  therefore  place  in  the  root. 

Now  the  root  may  be  represented  hj  T  -^  h-\-t  -{-  u,  T  standing 
for  the  thousands,  h  for  the  hundi-eds,  t  for  the  tens,  and  u  for  the 
units.     Hence  the  number  itself  will  be  represented  by  ( T  -\-h  -\-  t 

+  uf  =  T2  +  [2  r+  K]h  +  [2(  r  -f  A)  +  i]f  +  [2  ( r  +  /i  +  0  +  u]u. 

Having  removed  the  T2  (2  thousands  squared)  from  the  number, 
the  second  part  is  (27"-}-  ^K  the  lowest  order  in  which  is  hundreds 
multiphed  by  hundreds,  which  gives  ten  thousands  ;  wherefore  I 
bring  down  no  order  lower  than  ten  thousands,  or  simply  the  next 
period.  As  this  new  dividend  contains  (2  T  -|-  ^)K  I  will,  for  a  trial, 
consider  it  as  simply  containing  2  T  X  ^,  and  as  hundreds  into 
thousands  produce  only  orders  above  ten  thousands,  I  may  omit  the 
8  which  is  ten  thousands,  in  making  this  trial.  Using  2  T  or  4 
(thousands)  for  the  trial  divisor,  I  find  it  contained  in  32  (hundred 
thousands)  8  times.  But,  as  the  trial  divisor  is  too  small  by  this  new 
figure,  it  is  evident  that  adding  it,  thus  making  the  divisor  48,  it  will 
not  go  8  times.  Neither  will  it  go  7  times.  Thus  I  find  6  to  be  the 
next  figure  in  the  rootj  and  the  ti-ue  divisor,  IT  -\-  h,  to  he  46  {i.  e.  4 
thousands  and  6  hundreds).  Multiplying  4G  by  6,  and  thus  forming 
the  part  \^T -\-K]h,  I  find  it  to  be  276  (ten  thousands),  which  sub- 
tracted leaves  52  (ten  thousands). 

Again,  these  two  parts,  %dz. ;  Ta  and  (2  T*  -|-  h)h  having  been  re- 
moved, the  next  part  of  the  power  is  \2{T  -\-  h)  -f-  t]t,  or  [52 
hundreds  -j-  m-  As  the  lowest  order  of  this  part  is  tens  squared,  I  need 
bring  down  nothing  below  hundreds,  or  the  next  period.  (The  pupil 
can  now  fill  out  the  demonstration  as  in  the  preceding  paragraph.) 

[N.  B.  The  pupil  should  not  merely  extract  the  roots  of  the  follow- 
ing, but  he  should  apply  the  reasoning  as  above  given  in  full,  to  each 
example.  The  author's  experience  is  that  few  pupils  really  compre- 
hend the  reasons  for  the  processes  of  Evolution,  and  fewer  stiU  get  them 
fixed  in  the  mind.  ] 

2.  Extract  the  square  root  of  7225.  Boot,  85. 


3    to  7.    Show    that    \/i)80i   =    99,     v^iTU^y     =    217, 


\^553536  =  744,  \/4304G721  =  G561,  ^5704801  =  2401. 
8   to  11.      Show  that     v/."5=  .7071 +,  v/3    ^   1.73    +, 
v'50  =  7.071  +,  ^  =  2 .23G  +. 


EVOLUTION.  165 

210*  Cor. — In  extracting  the  roots  of  common  fi^actions, 
if  the  numerator  and  denominator  are  perfect  squares,  extract 
the  root  of  each  separately  ;  if  they  are  not,  it  is  usually  best 
to  convert  the  fraction  into  a  decimal  and  then  extract  its  root. 

.8164  +. 


211,  J^rob,  4,  To  extract  the  Cube  Boot  of  a  polynomial 
RULE. — 1st.  Aeeange  the  polynomiaij  with  reference  to 

ONE  OF  ITS  liETTERS,  AS  FOR  DIVISION. 

2nd.  Extract  the  cube  root  of  the  first  left  hand  teem. 
This  root  is  the  first  teem  of  the  required  root.  Subtract 
the  cube  of  this  teem  of  the  eoot  feom  the  polyno:miaii. 

3rd.  Take  three  tijies  the  square  of  the  boot  aleeady 

FOUND  FOE  A  TeIAL  DiVISOE.  By  THIS  TEIAIj  DIVISOR  DIVIDE  THE 
FIRST  TEEM  OF  THE  REMAINDER  OF  THE  POLYNOIMIAL,  AND  WRITE 
THE  QUOTIENT  AS  THE  SECOND  TEEM  OF  THE  ROOT. 

4tli.  Complete  the  divisor  by  adding  to  the  trial  divisor 
3  times  this  last  term  multiplied  by  the  part  of  the 
root  previously  found,  and  also  the  square  of  the  last 
term    found.      Multiply    the    true    divisor  thus    formed 

BY  the  last  term  IN  THE  EOOT,  AND  SUBTRACT  THE  PRODUCT  FROM 
THE  LAST  REMAINDER,  BRINGING  DOWN  SUCH  TERMS  AS  MAY  BE 
NECESSARY. 

5tli.  Take  three  times  the  square  of  the  root  already 

FOUND  AS  A  NEW  TRIAL  DiVISOR  ;  DIVIDE  THE  LAST  REMAINDER 
BY  IT  ;  COMPLETE  IT  BY  ADDING  TO  IT  THREE  TIMES  THIS  LAST 
TERM  MULTIPLIED  BY  THE  PAET  OF  THE  EOOT  PEE^TiOUSLY  FOUND, 
AND  ALSO  THE  SQUAEE  OF  THE  LAST  TEEM  FOUND  ;  MULTIPLY  AND 
SUBTEACT  AS  BEFOBE.  CONTINUE  TO  EEPEAT  THE  PEOCESS  OF 
FOEMING  TeIAL  DiVISORS,  DIVIDING,  COMPLETING  THE  DIVISOR, 
MULTIPLYING  AND  SUBTRACTING,  TILL  THE  POLYNOMIAL  IS  EX- 
HAUSTED, OR  UNTIL  THEEE  IS  NO  TEEM  OF  IT  EEMAINING  WHICH 
CAN   BE  EXACTLY   DTV^IDED   BY   THE    FIRST    TERM    OF     THE     TrIAL 

Divisor. 


166  EVOLUTIO]?^-. 

Dem.— 1st.  The  polynomial  is  arranged  as  in  division,  since  such 
is  the  order  which  the  terms  assume  in  cubing  any  polynomial,  as  the 
root  of  the  given  one  similarly  arranged. 

2nd.  In  cubing  any  polynomial,  the  first  term  of  the  cube  is  found 
to  be  the  cube  of  the  first  term  of  the  root ;  hence,  in  extracting  the 
cube  root,  the  cube  root  of  this  term  is  the  first  term  of  the  root. 

3d.  To  prove  the  process  of  finding  the  divisors  and  subsequent 
terms  of  the  root,  we  observe  the  following  operations  : 

(1)  d) 

A.  (rt  +  h)-'  =  a]  4-  3a^b  +  Sah^  -f  h''  =a^  4-[3a2  -f  3ah  -{-inb. 

B.  (a  +  ?>  +  cy  =  [(a  +  &)  +  c]3  = 

(a  J^  6)3  _|_  [3(a  _f  6)2  4.  3(a  +  h)c  +  c=^]c  = 

a''  _|_  [.3a!  -f  3ab-\-  62]?)  +  [3(a  +  h)''  +  S^t  +  h)c  +  c*]c. 

C.  (a  +  &  +  c  +  fZ)3  =  [^a  4-  6  +  c)  +  f7]^  = 

(a  +  6  +  cj^  +  [3(a  +  6  -f  c)*  +  3(a  +  6  +  c)d  +  d')d  = 

(1)  (2)  (3) 

a'  +  [3a2  +  dab  -]-h^]h  -{-  [3(a  +  &)«  +3(a  +  &)c  +  c^]c  + 
[3(a   -f  6  -f  c)'^  4-  3{a  +  h  +  c)d  +  d*]t/. 

Hence  it  appears  ;  1st,  That  the  cube  of  a  polynomial  is  made  up  of 
as  many  parts  as  there  are  terms  in  the  root ;  2nd,  that  the  first  part 
is  the  cube  of  the  first  term  of  the  root ;  3d,  That  the  second  part  is 
three  times  the  square  of  the  first  term  of  the  root  -{-  3  times  the  first  term 
into  the  second  term  -j-  the  square  of  the  second  term,  multiplied  hy  the 
second  ienn  of  the  root ;  4th,  That  any  one  of  the  parts  of  the  power, 
as  the  nth,  is  Three  times  the  square  of  the  n  —  1  preceding  terms  of  the 
root,  -{-  3  times  the  product  of  these  terms  into  the  next,  or  nth  term,  -f-  the 
square  of  this  last  or  nth  term,  all  these  terms  being  multiplied  hy  the  last, 
or  nth  term  of  the  root. 

Finally,  it  is  evident  that,  if  the  work  does  not  terminate  by  this 
process  when  the  letter  of  arrangement  disappears  from  the  remainder, 
it  can  never  terminate,  since  the  divisor  always  contains  this  letter. 

ScH.  1.  — If  the  first  term  of  the  arranged  polynomial  is  not  a  per- 
fect cube  the  root  cannot  be  extracted. 

ScH.  2.  — If  at  any  time  no  term  of  the  remainder  is  exactty  divisi- 
ble by  the  first  term  of  the  trial  divisor,  the  root  can  not  be  extracted. 


EVOLUTION. 


16? 


+ 


+ 


+ 


CO      , 
CO    + 


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s 

s  ^ 

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ei 

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1G8  POWEES   AND   ROOTS. 

Explanation — 1st.  I  arrange  this  polynomial  with  reference  to  a, 
and  thus  see  at  once  the  first  two  terms.  But  the  terms  36a'*c  and 
27a'*6c'^a;  are  of  the  same  degree  with  respect  to  a,  and  hence  to 
determine  which  is  to  have  tho  precedence,  I  notice  that  the  first 
term  in  the  root  mil  be  3«^  c,  and  as  the  second  term  of  the  polynomial 
divided  by  3  times  the  square  of  this  gives  the  second  term  of 
the  root,  I  observe  that  the  terms  containing  a  and  c  are  all  to  have^. 
precedence  over  those  containing  h  andic.  Hence  I  write  36a "^c  —  8a'^' 
next.  The  remaining  terms  I  arrange,  giving  a  the  precedence  and 
noticing  that  as  x  will  be  in  the  last  term  of  the  root,  its  higher 
powers  will  stand  last, 

2nd.  As  27a^c^  is  the  cube  of  the  first  term  of  the  root,  that 
term  is  3a^c,  which  I  consequently  place  in  the  root,  and  subtract 
the  term  27a^c'-^  from  the  polynomial. 

3rd.  As  the  second  i)art  of  a  cube  of  a  polynomial  is  3  times 
the  square  of  the  first  term  of  the  root,  plus  other  terms,  into  the 
second  term  of  the  root,  I  take  3  times  the  square  of  this  first 
term  of  the  root  or  27a "'c^,  for  a  trial  divisor.  Dividing,  I  find  the 
second  term  of  the  root  to  be  —  2a.  But  the  True  Divisor,  or  leading 
factor  in  this  second  part  of  the  power,  is  3  times  the  square  of  the 
former  part  of  the  root,  -f-  three  times  that  part  into  the  last  tenn 
found,  -}-  the  square  of  this  term.  Hence  I  add  3  times  3a^c  multi- 
plied by  —  2a,  and  —  2a  squared,  to  complete  the  divisor.  Having 
completed  it,  I  multiply  it  by  the  last  term  of  the  root  found,  —  2a, 
and  thus  form  the  second  part  of  the  power  of  the  root,  which  I  sub- 
tract from  the  given  polynomial. 

4th.  The  explanations  of  the  next  and  succeeding  steps,  when  there 
are  more,  are  identical  with  the  last,  and  can  be  supplied  by  the 
student. 

2.  Extract  the  cube  root  of  a^  —  85^  -\-  12a¥  —  Ga-b. 

Boot,  a  —  2b. 

3.  Extract  the  cube  root  of  5^;^  —  1  —  3^^  _|_  ^-c  —  3^7. 

Jloot,  x^  —  X  —  1. 

4.  Extract   the   cube   root  of   Q^Q^xA  +  1  —  GSj;^  —9^-1' 
8j7«  —  36.i;^  +  33^2.  .  Boot,  1x^  —  3:r  +  1. 

5.  Extract   the  cube   root   of   GOc^^^  +  48ca;5  —  Tic^  -f 
lOScs^  —  90c^^2  4.  8j^6  _  80c3j;3. 

Boot,  2x^  +  4cd;  —  3c2. 


I 


EVOLUTION. 


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170  POWERS  AND   EOOTS. 

2nd.   Take  the  cube  root  of  the  greatest  cube  in  the 

LEFT    HAND    PERIOD,  AND  WRITE  IT  AS  THE  HIGHEST    ORDER  IN  THE 

ROOT.  Subtract  the  cube  of  this  figure  from  the  period 

USED,  AND  TO  THE  REMAINDER  ANNEX  THE  NEXT  PERIOD  FOR  A 
NEW  DIVIDEND. 

3rd.  Take  three  times  the  square  of  the  root  already 

FOUND,  REGARDING  IT  AS  TENS,  FOR  A  TRIAL  DIVISOR,  BY  WHICH 
DIVIDE  THE  NEW  DIVIDEND.  ThE  QUOTIENT  IS  THE  NEXT 
FIGURE    IN    THE    ROOT  OR  A  GREATER    ONE.       To  OBTAIN    THE  TrUE 

Divisor  add  to  the  trial  dwisor  3  times  the  product  of  the 
last  root  figure  by  the  preceding  part  of  the  root,  regard- 
ed as  tens,  and  also  the  square  of  the  last  figure  in  the 
ROOT.     Multiply  this  true  divisor  by  the  last  root  figure 

AND  SUBTRACT  THE  PRODUCT  FROM  THE  LAST  NEW  DIVIDEND,  AND 
BRING  DOWN  THE  NEXT  PERIOD. 

4tli.  Take  three  times  the  square  of  the  root  already 

FOUND  REGARDED  AS  TENS,  FOR  A  TRIAL  DIVISOR,  AND  PROCEED  AS 
IN  THE  3rd  STEP.  KePEAT  this  PROCESS  TILL  ALL  THE  PERIODS 
HAVE  BEEN  BROUGHT  DOWN.  If  THE  NUMBER  IS  A  PERFECT  CUBE 
THE  REMAINDER  IS  ZERO.  If  IT  IS  NOT  A  PERFECT  CUBE  ANNEX 
PERIODS  OP  3  ZEROS  EACH  AND  CONTINUE  THE  OPERATION  TILL  THE 
REQUIRED  DEGREE  OF  ACCURACY  IS  ATTAINED,  MARKING  THE  FIGURES 
THUS  OBTAINED  AS  DECIMAL  FRACTIONS. 

ScH.  1. — In  pointing  off  decimal  fractions,  or  the  fractional  part 
of  mixed  numbers,  make  full  periods  of  three  figures  each,  annexing 
O's  if  necessary. 

ScH.  2. — If,  at  any  time,  the  trial  divisor  is  not  contained  in  the 
dividend  to  be  used  according  to  the  3rd  paragraph  in  the  rule, 
annex  a  0  to  the  root  and  also  two  zeros  to  the  trial  divisor,  bring 
down  the  next  period,  and  then  divide. 

ScH.  3.  —When  the  work  does  not  terminate  with  the  last  period  of 
significant  figures  it  will  not  terminate  at  all,  and  the  number  is  a 
surd.  This  is  evident  since  the  right  hand  figure  in  any  subtrahend 
arises  from  cubing  the  corresponding  digit  in  the  root,  and  the  cube  o/ 
no  digit  produces  0  in  unit's  place. 


EVOLUTION.  171 

Deu. — 1st.  That  this  method  of  pointing  gives  the  number  of 
figures  in  the  root  is  made  evident  by  cubing  a  few  numbers.  Thus 
the  cube  of  1  is  1,  and  of  10  is  1000  ;  hence  the  cubes  of  all  numbers 
between  1  and  10  have  1,  2,  or  3  (cannot  have  4)  figures.  The  cube  of  100 
is  1,000,000  ;  hence  the  cube  of  numbers  between  10  and  100  have 
4,  5,  or  6  figures,  but  can  not  have  7.  Again,  the  cube  of  1000 
is  1,000,000,000  ;  hence  the  cube  of  any  number  between  100  and 
1000,  i.  e. ,  of  any  number  expressed  by  3  figures,  contains  9,  or  one  or 
two  less  than  9  figures.  In  like  manner  it  appears  that  the  cube 
of  any  integral  number  contains  either  three  times  as  many  figures  as 
the  root,  or  one  or  two  less.  Since  in  multiphcation  of  decimal  fractions 
the  number  of  fractional  places  in  the  product  is  equal  to  the  number  in 
both  or  all  the  factors  used,  the  fractional  part  of  any  cube  must  have 
three  times  as  many  figures  as  the  root.  Thus  (2.15)3  _-  9.938375, 
six  fractional  places.  (.  612 )  ■*  =  .  229220928  or  nine  fractional  places. 

2nd.  That  the  greatest  cube  in  the  left  hand  period  is  the  cube  of 
the  highest  order  in  the  root,  appears  from  the  facts  that  the  cube  of 
any  number  of  units  between  1  and  9  falls  in  the  1st  period  ;  the 
cube  of  any  number  of  tens  between  1  and  9,  falls  in  the  second  ;  of 
any  number  of  hundreds,  in  the  3rd,  etc.  Moreover,  though  the  left 
hand  period  usually  contains  more  than  the  cube  of  the  digit  in  the 
highest  order  in  the  root,  it  can  not  contain  the  cube  of  a  unit  more  of 
that  order,  since  all  the  figures  that  can  follow  this  highest  order  in  the 
root  can  not  make  another  unit  of  that  order.      Thus    the   cube    of 

3999  can  not  be  as    great  as    the  cube  of  4000.     But  the   cube    of 

4000  gives  64  in  the  highest  period.    Hence  the  cube  of  3999  must 
give  less  than  64  in  that  period. 

3rd.  In  any  given  case,  supjoose  the  pointing  shows  that  the  root 
consists  of  thousands,  hundreds,  tens,  and  units.  Kepresent  the 
thousands  by  T,  the  hundreds  by  h,  the  tens  by  t,  and  the  units  by 
u.  Then  the  number  is  {T  -}- h  -}- t  -\-  u^  =  T'  -f  [3^^  +  327i 
+  h-'-]h+  [3(7+  h)'  +3(r+  h)t-i-t'-]t  -f  [3(r+  h  -\-t)^  4- 
3(1"  -j-  h -\-  t)u  -^u^]u.  Biit,  having  removed  the  cube  of  the 
thousands,  T^  the  next  part  of  the  power  is  [3  T^  -j- 3  Th -}- h^  ]h. 
No  part  of  this  can  fall  in  either  of  the  two  lowest  periods  of  the 
power,  since  its  lowest  order  arises  from  h'^  which  is  1,000,000,  at 
least.  Hence  we  need  only  bring  down  one  period.  For  a  irial,  we 
consider  this  part  as  31^  X  h,  and  hence  the  Trial  Divisor  is  BT'^ 
or  3  times  the  square  of  the  root  already  found.  Again,  regarding  this 
thousands  figure  as  tens,  makes  the  T,  which  squared  and  multipHed 
by  the  next  figure  of  the  root  which  is  also  hundreds,  give  millions, 


172  POWERS   AND   EOOTS. 

tlie  same  order  as  the  new  dividend.  But  the  True  Divisor  is  3T-  -J- 
337i  -f  A* ;  hence  we  add  to  3  T^ ,  3  T/i  -f  h^;  L  e.,  3  times  the  root  pre- 
viously found  multiplied  by  the  last  figure,  and  the  square  of  this  last  fig- 
ure. In  making  this  correction  we  are  to  remember  to  call  the  thou- 
sands so  many  tens,  which  reduces  it  to  hundreds,  the  order  of  the  root 
which  we  are  seeking;  whence  the  correction  becomes  the  square  of 
hundreds,  or  of  the  same  order  as  the  trial  divisor,  and  can  be  added 
to  it. 

dth.  It  is  evident  that  this  process  is  merely  repeated,  as  we  pro- 
ceed to  obtain  other  figures  in  the  root;  and,  as  the  law  of  notation  is 
the  same  as  we  pass  the  decimal  point,  no  special  exemplification  is 
needed  in  that  case. 

EXAMPLES. 

1.  Extract  the  cube  root  of  99252847. 

MODEL   SOLUTION. — OPERATION. 


G4 

T  3(40)2=  4800 

i  9.(A0\a  —     790 

Corrections 


_,  ht  u 
9925284714  6  8 


Trial  Divisor  3(40)2=  ^gOO 
3(40)6  =  720 
6^  =  ^ 
True  Divisor 5o5(i 


35252 


33336 


Trial  Divisor  3(460;=^  =  G3480U  1916847 
Corrections  I  ^(^g^l^i^^      ^'^^ 


True  Divisor 638949 


191 6847 


Explanation. — As  the  highest  order  in  this  number  is  ten  millions, 
the  highest  order  in  the  root  is  hundreds,  since  the  cube  of  a  hun- 
dreds figure  falls  in  millions  period,  while  the  cube  of  thousands  falls 
in  bilhons.  Moreover,  the  cube  of  the  hundreds  figure  is  the  greatest 
cube  contained  in  99,  i.  e.  64,  the  root  of  which  is  4,  which  is,  there- 
fore, the  hundreds  of  the  root.  That  the  hundreds  figure  is  not  greater 
than  4,  is  evident,  since  the  cube  of  5  hundreds  is  greater  than  the 
given  number.  But  the  cube  of  3  hundreds  with  the  greatest  pos- 
sible figures  in  tens  and  units'  places,  (i.  e.  399)  is  less  than  the  cub© 
of  4  hundreds.     Hence  the  hundreds  figure  is  not  less  than  4. 

Therefore  the  cube  root  of  the  given  number  is  A  -f-  f  -f-  w,  and  the 
number  itself  is  (A  -f-  f  -f-  w) »  =  K^  -f  [3h^  _f-  3ht  -f  i^]^  -f  [3(7i  -j-  t)^ 
-f-  (3/i  -f-  ^'^  -f-  w'^Jw.  But  having  removed  the  h-^  by  subtracting  the 
64  (millions),  the  next  part  of  the  power  is  [SA,*  -|-  3ht  -\-  f^lt.  Now  the 


EVOLUTION.  173 

lowest  order  of  this  is  i'\  or  the  cube  of  tens,  which  cannot  fall  below 
thousands,  so  that  I  need  only  bring  down  thousands  period,  i.  e.  the 
next  lower.  For  a  Trial  I  now  consider  this  part  of  the  power 
(35252)  as  3/1^  x  t,  or  3[40]2  X  ^  I  reduce  the  4  hundreds  to  the 
same  order  as  the  root  figure  which  I  am  seeking,  so  that  the  product 
of  its  square  by  this  root  figure  shall  be  of  the  same  order  as  the  new 
dividend.  Thus  my  new  dividend  is  thousands,  and  in  order  that 
thousands  divided  should  give  tens,  the  divisor  must  be  hundreds,  or 
the  square  of  tens.  Therefore,  reducing  the  4  hundreds  to  tens  it  be- 
comes 40,  whence  3(40)^  =  4800,  which  being  hundreds,  goes  into  the 
new  dividend,  which  is  thousands,  tens  times.  This  trial  divisor  is 
really  contained  in  the  dividend  7  {tens)  times,  but  as  the  corrections 
to  be  made  upon  it  for  the  true  divisor  are  so  great,  the  true  divisor 
does  not  go  but  6  times,  as  I  find  by  trying  7  for  the  tens  of  the  root. 
Having  thus  found  6  to  be  the  tens  of  the  root,  I  correct  my  trial  divi- 
sor, which  by  the  formula  is  3/i,*  -^3h  X  t  -\- l^,  hy  adding  Sh  X  t  or 
3(40)  X  6,  and  t^  or  6*,  and  find  the  true  divisor  to  be  5556  (hun- 
dreds). This  multiplied  by  the  6  (tens)  gives  the  second  part  of  the 
power,  i.  e.  {3h^  -f-  3ht  +  1^)1  =  33336  (thousands),  which  I  therefore 
subtract  from  the  given  number. 

[The  next  step  is  exactly  like  the  last,  and  the  pupil  can  supply  the 
demonstration.  But  be  sure  that  it  is  done,  and  the  reasoning  repeated 
in  subsequent  examples  till  it  is  familiar  and  fixed  in  the  mind.  ] 

2.  What  is  the  cube  root  of  74088?  Ans.  42. 

3.  What  is  the  cube  root  of  12326391  ?  Ans.  231. 

4.  What  is  the  cube  root  of  122P97755681  ?       Ans.  4961. 

5.  Wliat  is  the  cube  root  of  2936.493568?       Ans.  14.32. 

6.  What  is  the  cube  root  of  12.5  ?       Am.  2.321  nearly. 

7.  What  is  the  cube  root  of  .64?  Ans.  .8617  +. 

8.  What  is  the  cube  root  of  .08?  Ans.  .4308  -f. 

9.  What  is  the  cube  root  of  .008  ?  Arts.  .2. 

10  to  12.   Show  that  ^^2  =  1.2599  +  ;     ^5  =  1.7099  +  ; 

^9  =  2.08008  +. 
13  to  15.    Show  that  ^1  =  .87  +  ;  '^^^  =  |f  ;  ^34ti 


174  POWEES  AND  ROOTS. 

[Note. — The  author  does  not  think  it  worth  while  for  the  student  to 
consume  his  time  learning  the  various  shorter  methods  for  extracting 
roots,  the  various  methods  of  approximation  and  the  like,  inasmuch 
as  no  mathematician  thinks  of  using  them,  or  even  those  here  given, 
but  resorts  at  once  to  the  table  of  logarithms.  It  is  therefore  thought 
better  that  the  student  should  spend  his  time  in  becoming  perfectly 
familiar  with  the  demonstration  of  a  single  method,  than  to  cumber  the 
memory  with  a  multiplicity  of  processes  which  he  will  not  remember, 
and  which  if  he  were  to  remember  he  would  never  use.  ] 


FOR  REVIEW  OR  ADVANCED  COURSE. 

213,  JPvob,  G»  To  extract  roots  whose  indices  are  com- 
posed of  factors  of  2  and  3. 

Solution. — To  extract  the  4th  root,  extract  the  square  root  of  the 
square  root.  Since  the  4th  root  is  one  of  the  4  equal  factors  into 
which  a  number  is  conceived  to  be  resolved,  if  we  first  resolve  a  num- 
ber into  2  equal  factors  (that  is,  extract  the  square  root)  and  then  re- 
solve one  of  these  factors  into  2  equal  factors  (that  is,  extract  its  square 
root)  one  of  the  last  factors  is  one  of  the  4  equal  factors  which  com- 
pose the  original  number,  and  hence  the  4th  root.  In  like  manner  the 
6th  root  is  the  cube  root  of  the  square  root,  etc. 

EXAMPLES. 

1.  What  is   tlie   4tli  root  of  16a' —  QQa'x -{- 21Ga"-x^ — 
216ax^ -{- 81x'?  Ans.  ±  {2a— Sx). 

2.  What  is  the  6th  root  of  15:^2  —  20^^  ^  x^  —  6x'- -\- 1 — 
Gx  -f  15:^4?  Ans.  ±  {x  —  1). 

3.  What  is  the  6th  root  of  2985984? 

4.  What  is  the  8th  root  of  1679616? 


214,  JProb,  7.  To  extr^act  the  mth  {any)  root  of  a  deci- 
mal number. 

Solution. — Any  root  can  be  extracted  by  a  process  altogether  simi- 
lar to  those  given  for  the  square  and  cube  roots,  or  by  a  simple  inspec- 
tion of  the  corresponding  power  of  a  binomial.  Thus  to  extract  the 
fifth  root,  point  oif  by  placing  a  point  over  units  and  every  fifth  figure 


EVOLUTIOK.  175 

therefrom,  for  the  7th  root  over  every  7th  figure,  etc.  Extract  the  re- 
quired root  of  the  largest  power  of  the  mth  degree  in  the  left  hand  period 
for  the  first  figure  in  the  root.  Subtract,  and  bring  down  the  next  pe- 
riod. To  form  the  trial  and  true  divisors,  and  hence  to  find  the  other 
figures  of  the  root,  consider  the  corresponding  power  of  a  binomial.  Thus 
for  the  5th  root,  we  have  {a  +  b)^=a^  +  5a*b  +  10a3b^  +  10a'^b^  +  5ab*  +  b^^a^  + 
[5a*  +  10a^b  +  10aH'^  +  5ab^]b.  The  trial  divisor  is  5aS  i.  e.  five  times  the 
4th  power  of  the  root  already  found  regarded  as  tens.  The  corrections 
are  lOa^b  +  10a'^b^  +  5ab^ ,  regarding  a  as  the  root  already  found  and  as 
tens,  and  b  as  the  next  figure,  i.  e.  the  one  sought  by  the  trial. 

In  the  7th  root  the  trial  divisor  is  7a®,  and  the  corrections  are  21a56 
+  Z5a*b^  +  Zoa^b^  +  2'la^*  +  7ab^  +  b\ 

But  in  these  cases,  and  much  more  in  the  case  of  higher  roots,  the 
trial  divisor  differs  so  much  from,  the  true  divisor  that  the  process  is 
little  better  than  guess-work. 


2 IS,  I*Toh,  8,  To  extract  the  mth  root  of  a  polynomial. 

HULK, — Having  aekanged  the  polynomial  as  for  division, 
take  the  eoot  of  the  first  term,  for  the  first  term  of  the 
required  root. 

Subtract  the  power  from  the  given  quantity,  and  divide 
the  first  term  of  the  remainder  by  the  first  teem  of  the 

ROOT    IN\"0L'VT:D    to    the    next   inferior   POWER,  AND    MULTIPLIED 

by  the  index  of  the  given  power  ;  the  quotient  will  be  the 
next  teem  of  the  root. 

Subtract  the  power  of  the  terms  already  found  from  the 
given  quantity,  and  using  the  same  divisor,  proceed  as  before. 

Dem. — This  rule  demonstrates  itself,  as  the  final  operation  consists 
in  involving  the  root  to  the  required  degree. 

ScH. — This  rule  may  also  be  used  for  decimal  numbers. 

EXAMPLES. 

1.  Find    the   fourth    root   of   16a*  —  96a^x  +  216a^x^'  — 
216ax^  +  Six*. 

2.  Find  the  fifth  root  of  x^  +  5x*  -f  lOx"^  +  10j;2  +  5j;  +  1. 


176 


POWERS  AND  ROOTS. 


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178  CALCULUS   OF   RADICALS. 

SECTION  III. 

CALCULUS  OF  RADICALS, 


Eeductions. 

217 •  ^rob,  1,  To  simphfij  a  radical  by  removing 
a  factor. 

B  ULE. — Resolve  the  number  under  the  radical  sign  into 

TWO  FACTORS,  ONE  OF  WHICH  SHALL  BE  A  PERFECT  POWER  OF  THE 
DEGREE  OF  THE  RADICAL.  ExTRACT  THE  REQUIRED  ROOT  OF  THIS 
FACTOR  AND  PLACE  IT  BEFORE  THE  RADICAL  SIGN  AS  A  COEFFICIENT 
TO  THE  OTHER  FACTOR  UNDER  THE  SIGN. 

Dem.— This  process  is  simply  an  application  of  Cor.  Art.  205, 
which  proves  that  the  product  of  the  roots  is  equal  to  the  root  of  the 
product.  Thus  V^HoH^  =  v/lGa''6'^  X  'Sab  =  VlEcFb^  X  '^Sab 
=  ^a^hV'dah,  since  x/48a°6'  is  the  square  root  of  the  product  of 
IGcf^^*  and  3a&,  and4r)f^6\/3a6  is  the  product  of  the  square  roots,  [Ke- 
peat  the  demonstration  oi{19:4r)  and  Coe.  Akt.  205,] 

EXAMPLES. 


1.  Reduce  to  its  simplest  form  ^ibda'b^c\ 
model  solution. 

Operation.  flb\)aUj^  =  fWaFJFy7ia(?  =  f^ToH^^  X  ^  Tac^ 
=  3a&f/7ac^ 

Expi^ANATioN.  '^TbdaJbH^  indicates  that  ISda'^b^c^  is  to  be  resolved 
into  3  equal  factors.  Therefore  I  first  separate  it  into  two  factors, 
27a^b'^  and  7ac^,  one  of  which  is  a  perfect  cube.  Now  I  resolve  each 
of  these  factors  into  3  equal  factors,  one  of  each  of  which,  multiplied 
together,  will  constitute  one  of  the  three  equal  factors  of  the  given 
number.  27a^b'^  =  Sab  X  Sab  X  dab,  and  7ac^  =  ^Tac^  X  I^Toc^ 
X  ^7ac'^.  Hence  3ab'^7ac^  is  one  of  the  3  equal  factors  of 
189a^6'^c2  ;  or  f'lb'Ja'b^c'  =  SabflacT 


2.  Reduce  v21a^x=  to  its  simplest  lorm.  

liesuU,  Sax^'^'Saj- 


REDUCTIONS.  179 


3.  Keduce  ^SOa^x^  to  its  simplest  form. 

4.  Eeduce  '^^^'Slox^y^  to  its  simplest  form. 

SuGOESTioN.— Kthe  factor  of  the  decimal  number  under  the  radical 
is  not  apparent,  it  can  readily  be  found  by  a  few  trials.  Thus  in  the 
5rd,  we  could  try  the  square  of  2,  or  4;  and  then  the  square  of  3,  or  9; 
vad  then  of  4,  or  16;  of  5,  or  25;  of  6,  or  36;  of  7,  or  49;  of  which  we 
should  find  16  to  be  the  greatest  square  factor.  We  need  not  tiy  farther 
:han  49,  since  this  is  more  than  ^  of  80,  and  no  larger  number  can  be 
1  factor. 


•  5.  Simplify  ^UH-dx^y^. 

Suggestion. — Try  the  square  numbers  from  4  upward  till  you  find 
he  required  factor.  But  a  Uttle  judgment  will  save  labor.  Thus,  we 
leed  not  try  4,  for  no  number  multiphed  by  4  gives  a  3  in  units'  place, 
^^or  a  like  reason  we  would  try  9,  but  not  16,  25  or  36.  Then  again 
ve  would  try  49,  but  not  64;  81,  but  not  100;  121,  but  not  144.  Fi- 
nally 169  meets  the  case,  and  we  have  */llHdx^y^  =  ^l&^x^y^  X  ^^y 
=  ^ISdx^f'x  ^Txy  =  13x-i/n/7x^ 


6.  Simplifyv/3i7ya^6a;7. 


7.  Simplify  v  "diSa'x^.  Besult,  2ax^  '6a'x*- 


8.  Simplify  ^72yU.r"'^'^"+^  Result,  ^x^y''^l{)y\ 


9.  SimpKfy    (1715a^"-"6«") 


1 


L  an  -' 


Result,  la''"b"':y5a-'\  or  la' 

10.  Simplify  (352a%lo)^  Result,  2ah^'v'u. 

ScH. — Of  course,  by  the  use  of  fractional  exponents,  all  the  factors 
of  such  monomials  may  be  written  separately,  as  in  (202) .    Thus,  the 

1  i  a  i     2.    4. 

result    in    Ex.    7,    may    be    wi-itten    (96)^a''a;*,    or  \^)'^aa^xx^,    or 

-L  f  A 

2ax(3)''a  x  ,  or,  by  taking  the  5th  root  of  the  product  instead  of  the 

JL 

product  of  the  5th  roots  of  the  last  3  factors,  2rtx(3a3x')^,  as  above. 
The  method  of  this  rule  is  usually  appHed  for  removing  factors  which 
can  be  expressed  without  fractional  exponents. 


11.  Simplify  v^a^  —  a-2x.  Result, 


Suggestions.  >/«■' — a^x='s/a^(a  —  x)'=\/d^  X  v^a — x=.a\^  a — x. 


180  CALCULUS  OF  RADICALS. 


12.  Simplify  \/a"'+"6".  Remit,  abva'". 


13.  Simplify  v{a-—b-){a-{-b).        Result,  (a  +  h)Va  —  b. 

14.  Simplify  Z^'o^xK  Result,  Ibx^^^. 

Suggestion. — "When  the  radical  has  a  coefficient,  the  factor  removed 
from  under  the  radical  sign  is  to  be  multiphed  into  this  coefficient. 


15.  Simplify  {x  +  y)  v^^<  —  2x^y  -\-  xy'^. 

Remit,  (jp2— ?/')v/^ 

1,6.  Simplify  xyVx^y^  —  x^y\  Result,  x'^y"'\^x  —  y. 


17.  Simplify  Vx^^y-''z'"^+^.  Result,  x-^y-'^z'"^z. 

18.  Reduce  v^f  to  its  simplest  form.  Result,  ^^i>. 

ScH. — A  surd  fraction  is  conceived  to  be  in  its  simplest  form  when 
the  smallest  possible  whole  number  is  left  under  the  radical  sign.  The 
reason  for  this  is,  not  only  that  the  radical  factor  is  thus  made  simpler, 
but,  if  a  fraction  were  to  be  left  under  the  radical  sign  the  question 
would  arise,  What  fraction  ?    Certainly  not  the  least  possible,  for  such 

i2      ( — r 

a  fi-action  can  be  diminished  at  pleasure.     Thus  /(.|t^=/.|-1x'«  ^^ 

2     Itt  =  ^x  'j^6  X  q?"  "^^v/Mfi"'  ^*°''  ^^^•'  "^itliout  limit.     Perhaps, 

if  a  fraction  is  to  be  left  under  a  radical  sign  it  ^\ill  be  proper  to  con- 
sider the  expression  as  simplest  when  the  fraction  is  nearest  unity  ; 

|2  .  .  IT 

whence      pr  is  to  be  considered  as  simpler  than  8     1^^-. 

218.  Cor. — The  denominator  of  a  surd  fraction  can 
alivayf<  be  removed  from  under  a  radical  sign  by  multiplying 
both  terms  of  the  fraction  by  some  factor  ivhich  will  make  the 
denominator  a  perfect  power  of  the  degree  required. 

19.  Reduce  >J-,  >^-  ,^-  and  ^'-  to  their  simplest  forms. 


REDUCTIONS.  181 

ScH. — Thef  root  of  a  fraction  having  1  for  its  numerator  is  equal  to 
the  same  fraction  into  the  root  of  the  next  lower  power  of  the  denom- 


inator 


20.  Reduce/^  —  to  its  simplest  form. 

r  to  its  simplest  form. 

a  -\-  x-^ 

Result, "^a^  —  x\ 

a  -\-  X 

22.  Reduce  — -,   -,  -v  t^  and  4v  --—   to    their    simplest 

11\7   aWo  y   bb 

forms. 

BesuUs  (not  in  order),  •— v/l5^,  -Vg^,    and  — ^• 
dU  0  77 

23.  Reduce  7..v/ — —  to  its  simplest  form. 

a 
Result,  -r- —j-V^. 

24.  Reduce  /]-,  and  /Jj  to  their  simplest  forms. 

ScH. — The  above  simplifications  can  always  be  effected  upon  frac' 
tions,  but  only  upon  integral  radicals  when  the  integer  has  a  factoi^ 
which  is  a  perfect  power  of  the  degree  of  the  radical. 


219,  J^rob,  2,  To  simplify  a  radical,  or  reduce  it  to  its 
lowest  terms,  when  the  index  is  a  comjMsite  number,  and  the 
number  under  the  radical  sign  is  a  joerfect  power  of  the  degree 
indicated  by  one  of  the  factors  of  the  index, 

R  ULE, — Extract  that  root  of  the  number  which  corres- 
ponds TO  ONE  OF  THE  FACTORS  OF  THE  INDEX,  AND  WRITE  THIS 
root  as  a  surd  of  the  degree  of  THE  OTHER  FACTOR  OF  THE 
GIVEN  INDEX. 


182  CALCULUS  OF  RADICAI^. 

Dem. — Since  the  4tli  root  of  a  number  is  one  of  the  four  equal  factors 
of  that  number,  if  we  resolve  the  number  into  2  equal  factors,  and 
then  one  of  these  factors  into  2  other  equal  factors,  one  of  the  latter 
is  one  of  the  4  equal  factors  which  compose  the  given  number.  So, 
in  general,  the  ?nnth  root  is  one  of  the  mn  equal  factors  of  a  number. 
K,  now,  the  number  is  resolved  first  into  m  equal  factors,  and  then 
one  of  these  m  factors  is  again  resolved  into  n  other  equal  factors,  one 
of  the  latter  is  the  mnth  root  of  the  number. 

EXAMPIiES. 

1.  Eeduce  V25a^66. 

MODEL   SOLUTION. 

Opeeation.  V25a^— V  V'^do^P  =  ^Ba'^b*  =  ab'/'^b. 

Explanation. — The  4th  root  of  25a"*  6^  is  one  of  the  4  equal  factors  of 
it.  Hence  I  first  resolve  it  into  two  equal  factors,  one  of  which  is 
5a*  6^.  Then  I  resolve  5a- 6^  into  two  equal  factors,  or  rather  indicate 
it,  as  the  operation  cannot  be  fuDy  performed,  and  have  -J^a'^b^. 
■v/So^is,  therefore,  one  of  the  4  equal  factors  of  Iha^h^.  ^u<^'  ^y 
the  last  problem,  >/5a'^b^  =  ab\^5b.     Hence  V25a^6-*  =  abVUb. 


2.   Reduce  ^'21a'¥.  Result,  b\^^ab. 

Suggestions.   ^27a'b^  =  ^■1/270^=  ->/3a6*  =  b^Sab. 


3.  Eeduce  V~  64a^.  •      Besult,  2^—  a. 

4.  Reduce  v/25Ga'2j78. 

5.  Reduce  vSln'^m^^. 

6.  Reduce  \/x-^  —  2xy  +  y^. 


■  -  220*  JPvob,  S,  To  reduce  any  number  to  the  form  of  a 
radical  of  a  given  degree. 

RULE. — Raise  the  number  to  a  power  of  the  same  degree 

AS  THE  RADICAL,  AND  PLACE  THIS  POWER  UNDER  THE  RADICAL  SIGN 
WITH  THE  REQUIRED  INDEX,  OR  INDICATE  THE  SAME  THING  BY  A 
FRACTIONAL  EXPONENT. 


KEDUCTIOKS.  183 

Dem. — That  this  process  does  not  change  the  value  of  the  expression 
h  evident,  since  the  number  is  first  involved  to  a  given  power,  and  then 
the  corresponding  root  of  this  power  is  indicated,  the  latter,  or  indi- 
cated operation,  being  just  the  reverse  of  the  former.  Thus,  x  = 
\/x^.  That  is,  raising  x  to  the  mth  power,  and  then  indicating  the 
mth  root,  le&'7es  the  value  represented  unchanged. 

EXAMPLES. 

1.  Eeduce  la^x^  to  a  form  of  a  radical  of  the  3rd  degree. 

MODEL  SOLUTION. 

Operation.  7a'x^  =  Z^(Ja'^x'-^)*  =  ^34ya''x^. 

Explanation. — If  I  cube  7a*  x^  and  then  extract  the  cube  root  of 
this  cube,  the  result  will  evidently  be  the  same  as  at  first.  Now 
(7a^ic^)^  =  o-i:3a^x^.  But  instead  of  performing  the  operation  of  ex- 
tracting the  cube  root  of  SiSa^x^,  which  would  evidently  return  it  to 
7a^x^,  I  simply  indicate  the  operation,  and  have  '^'d'i^a^x^. 

2.  Reduce  2ay  —  3  to  the  form  of  a  radical  of  the  second 
degree.  Result,  ^4:a'y^  —  12a?/  +  9. 

3.  Reduce  a  —  :r  to  the  form  of  the  cube  root. 

|2 

4.  Reduce  /O-  to  the  form  of  the  4th  root. 


3 

5.   Reduce  -  to  the  form  of  the  3rd  root. 
5 


x^ 
6.  Reduce  -  to  the  form  of  the  4th  root. 
o 


Result 


'  ^i-i-o 


221.  CoR. — To  introduce  the  coefficient  of  a  radical  under 
the  radical  sign,  it  is  necessary  to  raise  it  to  a  power  of  the 


18-1  CALCULUS   OF   RADICALS. 

same  degree  as  the  radical;  for  the  coefficient  being  re- 
duced to  the  same  form  as  the  radical  by  the  last  ride,  we  have 
the  product  of  two  like  roots,  which  is  equal  to  the  root  of  the 
product  {194^  and  205). 

EXAMPLES. 

1.  Introduce  the  coefficient  in  3^ '^2^'  under  the  radical 
sign. 

MODEL.  SOLUTION. 

Operation.  Sxf^Sx"*  =  i^27aF  X  ^^^  =  1^27x^O<~2^  =  #^5^5^ 

Explanation. — Cubing  3x  I  have  27ic^,  the  indicated  cube  root  of 
which  is  F  27x"*.     This  is  evidently  the  same  in  value  as  3x.     Hence 

^xf^^  =  t^27x"^  X  ^2x*7  But  f^2nx^~X^  =  fWix^X  1^2x* 
since  the  former  indicates  one  of  the  three  equal  factors  of  27x^  X  2x*, 
or  54x^,  which  may  be  found  by  resolving  each  of  its  two  factors  27x* 
and  2x^  into  3  equal  factors  and  taking  one  from  each  group.  This, 
however,  is  what  is  indicated  by  l/27x^  X  1^2x^.  Hence  3x1^ 2x'^  = 
^27x^  X   1^^^  =    ^27x3  X  2x2    =  l^six^ 

[Note. — Doubtless  some  will  think  that  the  argument  contained  in 
the  above,  especially  in  so  elementary  a  form,  is  unnecessarily  re- 
peated. But  the  author's  experience  leads  him  to  think  that  such  a 
thing  is  scarcely  possible.  There  are  so  very  few  pupils  who  get  the 
idea  that  all  the  processes  of  Evolution  and  Involution  are  but  exten- 
sions and  applications  of  the  simple  principles  of  factoring,  that  it  is 
thought  best  to  make  use  of  every  opportunity  to  fix  this  fundamental 
thought,  and  elucidate  it.  Many  pupils  go  through  with  the  forms 
of  demonstration  usually  given  in  the  Calculus  of  Radicals,  and  really 
get  no  meaning  out  of  them.  ] 

2.  Introduce   the    coefficient    in   -\/2  under  the  radical 
sign;  inlv/ij;ini;j4;my9. 

Result.,  ^\,  Ji,  ^i,  Ji 


REDUCTIONS.  185 

3.  Introduce  the  coefficients  in  the  following  expressions 
under   the    radical    signs;    ^a'^'^'lax,  {a — x)'^ a  +  x, 


iv/4a,  {x  —  y)^x__  y^  and  -  ^27c 

o 


Two  of  the  results  are  V '^a^  —  x^){a  — x),  and  ^{^x—ij)\ 

4.  Introduce  under  the  radical  signs  the  coefficient  in 

q        j~q  

the  following:        ^^x\    o"  sJi^T*  5x\/25.r-',    and 


a  -\-h    \a^ 
a  —  h\a 


The  last  two  are  V5"»  +  2^"»-2^  and  \j 


222,  I^vob*  4. — To  reduce  radicals  of  different  degrees 
to  equivaloit  ones  having  a  common  index. 

BULE.—Bxthb&s  the  numbers  by  means  op  fractional 
INDICES.  Reduce  the  indices  to  a  common  denominator. 
Perform  upon  the  numbers  the  operations  represented  by 
the  numerators,  and  indicate  the  operation  signified  by 
the  denominator.  ^ 

Dem. — The  only  point  in  this  rule  needing  further  demonstration  is, 
that  multiplying  numerator  and  denominator  of  a  fractional  index  by 
the  same  number  does  not  change  the  value  of  the  expression,  i.  e.,  that 

a  ma  a^ 

a;6_ajOTb.  Now,  x'^  signifies  the  product  of  a  of  the  b  equal  factors 
into  which  x  is  conceived  to  be  resolved.  If  we  now  resolve  each  of 
these  b  equal  factors  into  m  equal  factors,  a  of  them  will  include  ma  of 
the  mb  equal  factors  into  which  x  is  conceived  to  be  resolved.  Hence 
ma  of  the  7nb  equal  factors  of  x  equals  a  of  the  b  equal  factors. 

[The  student  should  notice  the  analogy  between  this  explanation  and 
that  usually  given  in  Arithmetic  for  reducing  fractions  to  equivalent 
ones  having  a  common  denominator.     It  is  not  identical.^ 

EXAMPLES. 


1.  Reduce  s/ta'^x  and  ^J^:mHj  to  forms  having  a  conamou 
index. 


186  CALCULUS   OF   RADICALS. 


MODEL    SOLUTION. 


,2,A^ 


Opebation.       \/2a^x  =  (2a^x)' ,    and     '^im'^y  =  {4,m^yy  .      But 

{2a^xf  =  {2a^xf  =  (Sa^x^)^  =  v'Sa^;    and   {4:m^yf=  (4m  *yf  == 
s/lGm^y'^- 

Explanation. — Expressing  the  given  numbers  with  fractional  in- 
1  J. 

dices,  I  have  (2a^r)    and  {Am^T/)  .     These  indices  reduced  to  forms  hav- 

L 

ing  a  common  denominator  are  |  and  f.     Now  (2a^.v)    signifies  one  of 

3. 

the  ^100  equal  factors  of  2a  ^x  ;  while  (2a^x)*'  signifies  three  of  the  six 

J.  3.  '  i 

equal  factors  of  2a^x.  Hence  (2a'''x)  *=  (2a^x)  ^.     In  like  manner  {^m^y) 

I 
signifies  one  of  the  three  equal  factors  of  4m^^  ;  while  (4m* ^z)    signifies 

J.  2. 

two  of  the  six  equal  factors  of  the  same.     Hence  (4m^t/)^  =  (4m*2/)  • 
Finally,  as  (2a^x) ''  is  the  same  as  the  6th  root  of  the  cube  of  2a^x,  I 


have\/8a^x^.     In  like  manner  {■im^y)*',  meaning  the  6th  root  of  the 
square  of  A^nx^y,  becomes  s/l'om^y'^. 

[Note. — Let  the  pupil  show  that  the  sixth  root  of  the  square  is  the 
same  as  the  square  of  the  sixth  root,  etc.  ] 

2.  Reduce  V'^'and  v^to  forms  having  a  common  index. 

Residts,  ^/^^^^  v'a 

3  Reduce  VW,  ^/b',  \/2~to  forms  having  a  common  index. 

4  Reduce  ^^2^"^  ^^^,  and  ^/xyio  forms  having  a  common 
index.  One  result  is  V  729.'c«. 

5  Reduce  ^ax  and  's/hocP'  to  forms  having  a  common  index. 

Results,  (a^x^)^  and  {b^x')^- 

6  Reduce  «s  (5^*)^  and  (3c)«  to  forms  having  a  common 
index.  Results, ''^/2f(F/\/a^,  '^^/(j25b'. 

7  Reduce  x'"'  and  y"  to  forms  having  a  common  index. 


REDUCTIONS.  187 

8.  Eediice  2c^/x  and  ba\^'2y  to  forms  having  a  common 
index. 

Suggestion. — The  radical  factors  can  be  reduced  to  forms  having  a 
common  index  without  affecting  the  coeflScient,  since  the  operation 
does  not  affect  the  value  of  the  radical. 

9.  Kednce  4:^/bx■^y,  "H^^^xy,  and  lOaVSbx  to  forms  having  a 
common  index.     Results,  "l^/^xy,  \^a^%lh^x^,  4o^25x^y'^. 

10.  Eeduce  a+c  and  {a—cY  to  forms  having  a  common  in- 
dex. Results,  (a2  +  2ac+c2)2  and  Vct—c, 


223,  J^vob,  5,  To  reduce  a  /inaction  having  a  mono- 
mial radical  denominator,  or  a  monomial  radical  factor  in  its 
denominator,  to  a  form  having  a  rational  denominator. 

RULE. — Multiply  both  teems  of  the  fraction  by  the 

RADICAL  IN  THE    DENOMINATOR  WITH    AN   INDEX    WHICH    ADDED    TO 
THE  GIVEN  INDEX  MAKES  IT  INTEGRAL. 

Dem. — Since  two  factors  consisting  of  the  same  quantity  affected 
by  the  same  or  different  exponents  are  multiplied  by  adding  the  ex- 
ponents [90),  and  the  sum  of  the  exponent  of  the  denominator  and 
the  factor  by  which  we  multiply  it  is  an  integer,  the  product  becomes 
rational.  The  value  of  the  fraction  is  not  altered,  since  both  its  terms 
are  multiplied  by  the  same  number. 

EXAMPLES. 

1.  Reduce  —  to  a  form  having  a  rational  denominator. 

V2,x 

MODEL   SOLUTION. 

«.>x..,.^^x-      5^-^        3a(2a;)2  X  (3a;)2       3a(6as2)^  .rr 

OPEBATION.       -iznr-  = '. i = ..  =a\/^. 

^^*  (3x)*  X  (3a;)«  '^"^ 

Explanation. — Using  fractional  exponents,  I  have  — _— .      Since 

(3x)^ 
mtdtiplying  numerator  and  denominator  by  the  same  number  does 

not  alter  the  value  of  the  fraction,  and  as  (3x)'^  X  (3x)^  makes  3a-,  I  can 


188  CALCULUS   OF  BADICALS. 

rationalize  the  denominator  of  this  fraction  by  multiplying  both  its 

terms  by  (3x)*      This  gives -^-; which  reduces  to  a\/6.      [K  the 

iiX 

rationale  of  these  last  reductions  is  not  perfectly  familiar  it  should  b© 

given.      Thus  (6a;-')    =  ^6)  ^  (x^^)  ^  =  6  ''x,  vrhence — —  =      ._,^    ,  and 

Cancelling  the  3x  I  have  un/B.] 

2 

2.  Rationalize  the  denominator  of  — -=, 

2  2  J 

Suggestion.     =r-  =  —^,  and  a^  is  the  factor  which  rationalizes  ii 

3v^a^       3a^ 

oa- 

3.  Rationalize  the  denominator  of  —_. 

4.  Rationalize  the  denominator  of  -~  •     Result, ^  . 

'va  a 

5.  Reduce  -;;^  to  a  form  having  a  rational  denominator. 

Eesult,  ^_^^. 


12        6  ^3 

6.  Reduce  —=,  --=.,  -t7=^,  and  —   to  forms    having  ra- 
v3    V3    V  6i  v6 

3    

tional  denominators.  One  of  the  results  is  kv/16. 

v/3~      (3)^X(6)^        ^ig-      3^2"       1 
Suggestion.      _ — = ; =  ^       =  Ll =  -  ,/.J" 

>/6"      (6r^x(6r  ^  ^  ^ 

ScH. — This  process  is  equally  applicable  to  any  form  of  radical  fae< 

ior  in  the  denominator,  whether  monomial  or  polynomial. 


7.  Rationalize  the  denominator  of 


Suggestion. 


y/g  — X       {a  —  xy^  x  (n -{- ^)^  _  s/n'^—x'^ 
A  +  iC       (a  -fx)^'  X  (a  +  x)^         "  +  '"^ 


REDUCTIONS.  189 

8.  Reduce —  to  a  form  having  a  rational  denom- 

inator.  Besult, ^ ^ 

6c  —  'Ax-^ 


2 2  J:*  IPvoh,  6,  To  rationalize  the  denominator  of  a 
fraction  when  it  consists  of  a  binomial,  one  or  both  of  whose 
terms  are  radicals  of  the  second  degree. 

RULE. — Multiply  both   tkrms   of  the  fraction  by  the 

DENOMTNATOK   WITH    ONE    OF   ITS    SIGNS    CHANGED. 

Dem. — This  rationalizes  the  denominator,  since  in  any  case  it  gives  the 
product  of  the  sum  and  difference  of  the  two  terms  of  the  denomina- 
tor, which  being  equal  to  the  difference  of  their  squares,  frees  either 
or  both  from  radicals,  as  the  square  of  a  square  root  is  rational 

EXAMPLES. 

n 

1.  Rationalize  the  denominator  of 


^b 


OPEEATION. 


MODEL   SOLUTION. 


a—Vh        t^a  —  ^bXai-^b)  a^  —  b 

ExpiiANATioN. — I  observe  that  a  —  ^b  will  be  rationahzed  by  mul- 
tiplying it  by  a  -f  Vb,  since  the  product  of  the  sum  and  difference 
of  two  quantities  is  the  difference  of  their  squares.  Hence  multiply- 
ing both  terms  of  the  fraction,  so  as  not  to  alter  its  value,  I  hava 
a-  -\-  a^^b 


2.  nationalize  the  denominator  of 


^a—^b 


190  CALCULUS   OF   RADICALS. 

3.  Reduce =:  to  a  form  having  a  rational  denom- 

inator. 

3(2v/2  +  3  \/3-)         6v/2  +  Ov^B 


Result, 


8  —  27  —  ly 

6\/2  +  9v/3 


g  5\/2 

4.    Reduce -^  to  a  form  having?  a  rational    de- 

3  —  2^2 

nominator.  Result,  4  +  \/% 

l—\/l      1    _^   \/2 


5.  Rationalize  the  denominators  of 


3  +  v/5      2    +   V'A 


3v/5  —  2^2        ,^a  +  x  —Va  —  x 

— =,  and 

2n^5  —  v^lS  \/a  +  ^  +  ^a—  X 


Results,  2  —  v/5,  iv/2,  9  +  |v/lO,  and -, 

X 

6.  Rationalize  the  denominator  of 


"^X-   -f   .77   -f   1    Vx'^  X  1 


Vx-  -Y  X  -\-l    +    Vx^-  —  X  —  1 


„     ,,   ^2  — ^y^i  —  ^2  —  2^ 

Result,   


^-  +  1 


FOR  REVIEW  OR  ADVANCED  COURSE. 

22S*  JProh,  7. — A  factor  may  he  found  which  will  ror^ 
tionalize  any  binomial  radical. 


Dem. — If  the  binomial  radical  is  of  the  form  /(«  -f-  by",  or  (a  +  &)" » 

n  —  m 

the  factor  is  (a  -f-  5)    "  ,   according  to  {223), 

i.        I.  i. 

If  the  binomial  is  of  the  form  V^a*    -j-  Z^W  or  «"*  -j-  Z»"      Let  a'"  = 


REDUCTIONS.  191 

1  1  *■ 

a;,  and  b"  =y;  whence  a'"  =  x' ,  and  h"  =y  .     Also  let  p  be  the  least 

common  multiple  of  m  and  7i,  whence  x"^  and  y*"^  are    rational.       But 

gp  rp 

x'P  =  a"*,  and  y*"^  =  &".     If  now  we  can  find  a  factor  which  will  ren- 

•  r 

der  xf  -\-  y'',  x'P  ±  y"^,  this  will  be  a  factor  which  wiU  render  a"*-)-  &"  , 
^m  _^  ^i»  which  is  rational.     To  find  the  factor  which  multiphed  by 

^'   +  2/"^   gives  x^  4-  y'"'',  we  have  only  to  divide  the  latter  by  the 

x'P  4-  yp 
former.     Now        —  ^    ^x^p-i)  — ay(p— *)y  -j-x»(p— ')y2''  —  x''(p-*)y5'"-(- 

±  y(p— 1)    (^),  the  -j-  sign  of  the  last  term  to  be  taken  when  p  is 

odd,  and  the  —  sign  when  it  is  even  {126).      Therefore  ic»0'—i)  — 
x*(p-2)y'"  -j-  a;»(p-')y2'"  —  x^(p—^'>y^''  -j •   4-  y''P— i),  is  a  factor  which 

will  render  v^a*  +  -^^ '"  rational,  x'  being  understood  to  be  a"*,  and 

r 

y  =  h^,  and  p  the  L.  C.  M.  of  m  and  n. 

If  the  binomial  is  ^a'  —  v^6'" ,  the  factor  is  found  in  a  similar  man- 
ner, and  is  x»(p-^>  -j-  x'^^^^^y  +  x'^p-^^y'^''  -| f-  y(»>-i). 

EXAMPLES. 

/—         a/-  J-  1 

1.  Kationalize  va  +'v6  or  a^  +  6^. 

i.  1 

Solution. — In  this  example  s=l,  r  =  1,  p  =  6,  x  =  a*  andy=6^. 
5.  J.         a  3.  -L  A        5. 

Hence  formula  (A.)  becomes  a"^  —  a'-b'^  -j-  a'^"*  —  ab  -^  a* 6'*  —  6"*. 

This  factor  multiphed  into  a^  -f-  ^^  gives  a^  —  &-,  as  the  rational- 
ized product. 

2.  Find  a  factor  which  will  rationalize   '^e^   —  v 


u3  or 


2         3 

22  20^  186.  16^  14      12 

_1_3    J_6  1_0_    1_8  8.    2Jl  6     2_4  4     2_7  2     3_0 

3_3 

i>  *  ,  is  the  required  factor,  and  the  expression  ration- 
alized is  f?^  —  t"'. 


192  CALCUXiUS  OF   RADICALS. 

226.  Proh.  8,  A  Trinomial  of  the  form  v^a  -|-  v^b  + 
v^c  may  he  transformed  into  an  expression  with  hut  one  radical 
term  hy  multiplying  it  hy  itself  with  one  of  the  signs  changed, 
as  v^a  +  ^b  —  v^c.  The  product  thus  arising  may  then  he 
treated  as  a  hinomial  radical  hy  considering  the  sum  of  the 
rational  terms  as  one  term,  and  the  radical  term  as  the  other. 

Thus,  {^a+  ^b+  v^c) (^a  +  A  —  ^c)  =  a  +  6  —  c  + 
2^ab.  Again,  [{a  +  h  —  c)  +  2^aF]  X  [(a-f  5  — c)  — 
2^ ah]  =  a-^  +  6^  +  c^  —  2ab  —  2bc  —  2ac,  a  rational 
result. 

Ex.   1.  Rationalize   v^S  —  v^l  _  \/3.  Besult,  4. 


SECTIOJsf  IV. 

COMBINATIONS  OF  RADICALS. 


ADDITION  AND  SUBTRACTION. 
227.  l?roh.  1.   To  add  or  subtract  radicals. 

RULE. — If  the  badigals  ake  similar  the  rules  already 

GIVEN  {72f  77)  ARE  SUFFICIENT.  If  THEY  ARE  NOT  SIMILAR 
MAKE  THEM  SO  BY  {217-222),  AND  COMBINE  AS  BEFORE.  If 
THEY  CANNOT  BE  MADE  SIMILAR,  THE  COMBINATIONS  CAN  ONLY  BE 
INDICATED  BY  CONNECTING  WITH  THE  PROPER  SIGNS. 

Dem. — When  the  radicals  are  similar  the  radical  factor  is  a  common 
quantity  and  the  coefficients  show  how  many  times  it  is  taken.  Hence 
the  sum,  or  difference,  of  the  coefficients,  as  the  case  may  be,  indicates 
how  many  times  the  common  quantity  is  to  be  taken  to  produce  the 
required  result. 

If  the  radicals  are  not  similar,  the  reductions  do  not  alter  their 
values ;  hence  the  sum  or  difference  of  the  reduced  radicals,  when 
they  can  be  made  similar,  is  the  sum  or  difference  of  the  radicals. 


COMBINATIONS — ADDITION  AND   SUBTRACTION.        193 
EXAMPLES. 

1.  Add  v/T6  and  ^212^ 

MODEL   SOLUTION. 

OPEBATION.     V  i»  =  3>/l:,  and  v^2'12    =   ll'v/2. 
.-.   VU  -f-   ^^2   =  3v/iZ  -f-  11^2  =  14^2. 

Explanation,  x/18  =  x/y  x  '^-  But  the  square  root  of  the  pro- 
duct equals  the  product  of  the  square  roots;  hence  ^^d  X  ^  =3V2.* 
In  hke  manner  x/24:2  =  x/l21  x  2  =  llv/2.  Therefore  V'lb  -f- 
'v/242  =  3V'2  4-  11  v/2.  But  three  times  any  quantity,  as  ^2, 
and  11  times  the  same  are  14  times  that  quantity.  . •.  3^2  -f-  H'^^ 
=  14^2. 


2.  Add  \^'24:oxy'  and  vj^^xy^.  Sum,  lly^'Sx, 

3.  Add  ^5UU  and  '^lOH. 

4  Add  v^a-?/  and  ^""c-y.  ^um,  (a  +  c)v/|/. 

5.  From  v/6(J5"take  the  ViOST  X>?/f.,  Sv^S. 

6.  Add  ^GU5  and  — v/i05. 

7.  Add  3^  -  and  2^  J- 

^N5  NflO. 

8.  From  3  J^  take  -2^^-  ^iff.,  |v/lO. 

9.  AVhat  is  the  sum  of  .    -  and  ;,.  I-?  ^/28.,  -v/3. 

10.  "WTiat    is    the    difference    of    ^2ax-  —  -ku.r  +  2a    and 
^y'lax-  -f-  4a j;  4-  i:a  ?  Ans.,  2^2a. 


*  If  this  reasoning  is  perfectly  familiar  it  may  do  to  omit  it.  But  it  is  far  better  to 
repeat  a  reason  when  it  is  already  clearly  comprehended,  than  to  omit  it  when  there 
is  the  least  doubt.  If  it  is  famiUar  it  can  be  given  in  an  instant ;  and  if  it  is  not  fami- 
liar it  ought  to  be  made  so  by  repetition  and  further  study. 


194  CALCULUS   OF   RADICALS. 

"Why  should  no  sign  be  given  to  the  last  answer  ?  If  the  problem 
read,  From  v/2ttx=^  —  4ax  +  2a  take  >/2ax^"+lax  +  2o,  why  would  the 
answer  be  — 2v/2a? 

11.  What  is  the  sum  of  {a  —  xy^xy  and  {a  +  xY^xyl 

^?2S.,2(a2  +  x'')^x.j. 

12.  What  is  the  difference  of  (a  —  x)^Vxy  and  {a^xy^xy  ? 

3      1 

Ans.,  ^ax'^y'^. 

13.  Find  the  sum  of  8  J5,  Vm,  —1\^Th,  and  A-. 

Sum,    4\/ij. 

14.  From  ^  I^Z^  take  ^llE      Bern. ,  {^^x -r,)  ^ ^. 

15.  Add  «Ji+r-i*  and  ^Ji+rf]*. 

Suggestions.      «     ll_l-r-|-'     =    a>Jl+—   =  a\j^^ -^  = 

^5Va^-f6^      lu  like    manner    ?>    jl  +  [-1 '     =  ^' \^  J  4.  ^^i 
.*.    The    sum    is    (a^  +  6^)  V  J^lTf^  =   (J  -{-b'){J -^b'Y^  == 


16.  From  (a  —  x) v^a-  —  :z;'  take  {a  — ■  x), 


'Sa 


X 


Bern.,    (a  —  x  —  l)v/a2 — x-\ 


17.  Add and   - 


X-^^^X'^ 1  X — "^x- 1 

Suggestions. — Beducing  the  given  fractions  to  a  common  denomi- 
nator, they  become  X — s/x- — ^1  and  x-j-s/x- — 1.  Hence  the  sum 
is  2x. 

18.  Add  -^^^n  +  v^.~i  ^^^  ^^^rjl_V7~y 

Vx''  +1  —  ^x-'-  —  1  Vx^  +  1  +  ^^•-  —  1 

Sum,  2;r2. 


COMBINATIONS — MULTIPLICATION.  195 

MULTIPLICATION. 

[Note. — Although  the  principles  embraced  in  this  section  have  been 
evolved  in  the  preceding  chaj)ters,  under  appropriate  heads,  their  im- 
portance is  so  great  that  it  is  thought  best  to  collect  them  here,  and  in 
connection  with  a  careful  review,  extend  somewhat  their  apphcation.  ] 

228,  I^rox>.  1,  The  product  of  the  same  root  of  two  or 
more  quantities^  equals  the  like  root  of  their  product. 

Dem. — Thatis^yx  X  \/y  =  \/-^y-  This  is  evident  from  the  fact 
that  ^xy  signifies  that  xy  is  to  be  resolved  into  m  equal  factors.  If 
now  each  factor,  as  x  and  y,  be  separately  resolved  into  in  equal  factors 
and  then  the  product  of  one  factor  from  each  be  taken,  there  -wdll 
be  m  such  equal  factors  in  xy.  Thus  ^x  is  one  of  the  m  equal  factors 
of  X,  and  ^y  is  one  of  the  m  equal  factors  of  t/.  Hence  [y^X  'Yy] 
X  \_\/x  X  s/y]  X  Lv^X  v^J/]  etc.,  torn  factors  of  ^.c  X  \/?/,  makes 
up  xy.     Therefore  \/x  X  s/y  =  \/xy.     (See  Arts  .  205  and  202, ) 


229,  I^rox>,  2.  Similar  Radicals  are  multiplied  by 
m^ultiplying  the  quantities  under  the  radical  sign  and  writing 
the  product  under  the  common  sign  ; 

Or  by  indicating  the  root  by  fractional  indices,  and,  for  the 
product,  taking  the  common  number  unth  an  index  equal  to  the 
sum  of  the  indices  of  the  factors. 

Dem.  1st. — Since  similar  radicals  are  the  same  root  of  the 
fame  quantities,  as  X^x  X  v^x,  this  is  only  a  partictdar  case  under 
Br  op,  1, 

1         1 

2nd.  x'"  X  X  "'  signifies  that  one  of  the  m  equal  factors  of  x  is  to  be 
multiplied  by  another  of  the  m  equal  factors,  or  by  itself.     This  gives 

2 

2  of  the  m  equal  factors  of  x,  which  is  what  is  indicated  by  x"*. 


230.  J^rob,  2.  To  multiply  radicals. 
RULE. — If  the   factors  have  not  the  same   index,  re- 
duce   THEM    TO    A     COMMON    INDEX,     AND     THEN    MULTIPLY    THE 


196  CALCULUS    OF   EADICALS. 

NUMBERS     UNDER     THE     RADICAL     SIGN    AND    WRITE     THE     PRODUCT? 
UNDER    THE    COMMON    SIGN. 

Dem. — (This  is  the  same  as  Prop,  1,) 

EXAMPLES. 

1.  What  is  the  product  of  ^^1  and  ^J) 

MODEL    SOLUTION. 

Opekation.  x/2=  VbTand  1^3  =  Vo.    .  • .  •v/2  X  1^3"=  Vs  X  t^  9 

Explanation.  V  2  =  ^/y,  since  the  former  is  one  of  the  two  equal 
factors  of  2,  aBcl  the  latter  is  three  of  the  six  equal  factors  of  2.  In  like 
manner  f'd=^'^.  Consequently,  >/2  X  1^3  =  v'8  X  ^9.  Now 
since  the  joroduct  of  the  same  root  of  two  numbers  is  equal  to  the  lik« 
root  of  the  product,  -v/s  X  4/9=  ^W. 

2.  Multiply  v^te  by  ^2aa  Prod.,  'y^2a^. 


3.  Multiply  v^a  —  xhj 


Prod.  Va^  —  ba'x  +  10a^.x-- —  lUa-x'^  +  ^ax^ —  xK 
4.Multiply  Jjby  JA.  p^od.,  i 

5.  Multiply  ya  by    ^^3. 

Prod,  3^"^^=  3"^^  =  ^6561. 


6.  Multiply  v''2a.r  by  v/2a^. 


Prod.,  (2ax)^^  =  y4096fti2j;i2 

7.  Multiply    ^^    by    ^| 

Pro6?.,  fi  -  =    -v^iyST 
\3  3 

8.  Multiply  3v/2a^by  2^^^. 


COMBINATIONS — MULTIPLICATION.  197 

SuG. — Here  we  have  the  continuous  product  of  3,  ^'lax,  2,  and  "^xy. 
But,  as  the  order  of  the  factors  is  immaterial,  we  may  write  3  X  2  X 

9.  Multiply  3>g?  by  6^i.  Pro^.,  6  v^236196. 

10.    Multiply    5a^  by  da^ 
IL    Multiply   2v/a6'by  3^a6! 

12.  Multiply  4.ah^hy5ah\ 

13.  Multiply  Sx^y^  by   2x^y^  and  express   the  product 
without  fractional  exponents. 

14.  Multiply.  I-  by  A-^  and  express  the  product  without 

the  use  of  the  radical  sign  and  in  its  simplest  form. 

Prod.,  Jl(9000)^. 
10 
i.         JL 

15.  Multiply  a"*  by  6". 

1  1  1 

Prod.,a''b'^  or,  Va"Z>'",  or  (a"6'")"'". 

16.  Multiply    iv^byi^iO.  Frod.,^t/250. 

17.  Multiply   a^^x,  h^y,  and  cv/J  together. 

Prod.,  abc^^x"^-  y'"''-  z^\ 

18.  Show  that  2^^  X  16  =  16vl2. 

19.  Show  that  ^24  X  6V3  =  6v/l2. 

20.  Multiply  2v/a  —  Z^bhy  Sv^a  +  2v^. 

OPERATION 

2v/a  —  3v/6 
3-v/a  4-  2x//7 

"^  6a      —  9^/06 

4-  4x/a^  —  65 

6a    —  by^ajb  —  66,  or  6(a — 6)  —  Sy oS! 


198  CALCULUS   OF  RADICALS. 

21.  Multiply  3  +  v/5  by  2  —  ^.  Prod.,  1  —  V^. 

22.  Multiply  v^  +  1  by  v/2  —  1.  Prod.,  1. 

23.  Multiply  IW2  —  WTb  by  v/(5  +  /a  _  _ 

Prod.,  2^3  —  v/lO. 


24.  Multiply  V 12  +  ^  ly  by  V 12  —  >/l9. 

SuG.  Since  these  two  radicals  are  of  the  same  degree  we  may  mul- 
tiply the  quantities  under  the  common  sign,  and  write  the  product 
under  the  same,  which  gives  v  144  —  19  =  5. 

25.  Multiply  a"-  —  av/2  4-  1  by  a-  +  ^^^2+1. 

Prod.,  a^  +  1. 

26.  Expand  {x^-  +  DC^^  —  ^^3  +  l){x^  +  ^v^3  +  1). 

Prod.,  x^  -\-  1. 

27.  Multiply  3^45  —  7^^  by  n/i|  +  2v'ij|.       Prof^.,  34. 

28.  Multiply  Va  +  c^6  by  v/fl  _  c^6. 

Proc?.,  a  —  d'-'^bK 


DIVISION. 


2St.,  ^rop.    The    quotient    of  the   same  root   of  two 

quantities  equals  the  like  root  of  their  quotient. 

Dem.  —Let  m  be  any  integer  and  x  and  y  any  numbers  ;  we  are  to 

to/-  I—  m  |— 

prove  that  ^yx-^^y,  or  ^  =  l^y-     Now,  that ^^  ^y 
dent,  since  ~—  raised  to  the  mth  power,  that  is 

yy 

V^  X  \/x  X  Z^x  X  v/a:  -  -  -  to  m  factors       x       ,  .. 

-^^  _        ^._ ^1-^ ^— =  -  ;  whence  it  appears 

\/y  y<  \/y  ^  '\/y  X  C^y  —  to  m  factors     y 


isevi- 


X 


that  - —  is  the  mth  root  of  -  or  equals  ^  r.       (See  Abts.   206  and 
y-y  2/  -' 

202,) 


\y 


COJVIBINATIONS — DIVISION.  199 

232,  Proh,  3.  To  divide  Radicals. 

R  ULE. — If  the  radicals  are  of  the  same  degree,  divide 

THE  NUMBER  UNDER  THE  SIGN  IN  THE  DIVIDEND  BY  THAT  UNDER 
THE  SIGN  IN  THE  DmSOR,  AND  AFFECT  THE  QUOTIENT  WITH  THE 
COMMON  RADICAL  SIGN. 

If  the  radicals  are  of  different    degrees,  reduce  THEM  TO 
IHE  SAME  DEGREE  BEFORE  DIVIDING. 

Dem. — [Same  as  above  ;  or,  it  may  be  considered  as  the  converse  of 
the  corresponding  case  in  multiplication.  ] 

EXAMPLES. 

1.  Divide  ^'6a-y^  by  ^2^. 

MODEL   SOLUTION. 

Opeeation.      v^3a-?/3  =  t/27a"y\  and  ^2ay  =  y^a-yK 
.     v/3^  _^^27^3_  J27^B_    J27^^_le,^,^    ,    , 
'   '    ^2^y  VT^      Nj^«^r       nJ    ^  2^43.a2/. 

Explanation. — Since  \/'da-y'^   =  \/27a^t/*,  and   ^'Aay  =  v^-ta'^j/S 

' rrrr-  '-=  '  :.     And  since  the  quotient  of  the  6th  root  of  two 

^2ay  V4a=y2  

^27a&j/9  (27aV 

numbers  is  the  6th  root  of  their  quotient,  v^  =f  \.^„  '   ,  which, 

V  40=2/2       \4a-?/2 

by  performing  the  operation  indicated,  becomes  .^  / — j —  ;  and  by  re- 
ducing so  that  the  number  under  the  radical  sign  shall  have  the  inte- 
gral form,  this  becomes  -  3/432a-?/7 

2.  Divide  ^/l^Ea^x^  by  \^5a^xy.  QuoL,  5v^. 

3.  Divide  v/3  by  ^.  Quot,  ^^=2'^^2 

4.  Divide  J?  by  WL  QuoL,  -Myii. 


y 


5.  Divide  '^2a-^x3  by  y2a^x^  QuoL,  s/Aa'x^. 

6.  Divide  ^72  by  v^.  '  QuoL,  V^'S. 


200  CALCULUS   OF   RADICALS. 

2     1  11 

7.  Divide  x'^y^  by  x'^y^,  and  express  the  quotient  without 

fractional  or  negative  exponents.  Quot,    « F. 

8.  Divide  24v^  by  Q^a^yK 

6^^  ^a^        'v'aY         "S'^'y"  y'^y 

9.  Divide  125^x'^y^  hy25'^x^y^,  expressing  the  quotient 
without  the  radical  sign.  QuoL,  5x^y^. 

10.  Divide  v^  by  ^4.  QuoL,  v^a^. 

11.  Divide  20^200  by  4^.  QmL,  5^5- 

12.  Divide   '  k  by    'k. 

13.  Divide  Jv^  by  v'2  +  Sv^I 

Suggestions.      s/^   -\-  S^/k   =  ^\^k  +  3\/4  =  5>/i      Whence 

Wi  ^vT      i       1 


^2  +  3^4        5\/4        5        10 

14.  Divide  ^a^  —  6-^  by  ^^TITft.  Quo/5.,  ^7:^^. 

15.  Divide  {a^b-^c)^hj  (ab)K  QuoL,  a'^bc. 

16.  Divide  200  by  v/io.  Q^o^.,  lOv'io,  or  (10)^. 

17.  Divide  a^^x  —  ^^bx-^-a'A/  —  v^by  by  ^x  +  v^t/. 

Suggestion. — Observe    that    a'^x  —    '>/bx'-\-    a-s/y  —    ■\/6y"  = 

a(  Vic  +  Vy)  —  x/6  (Vx  +  Vj/ )  =  (a  —  v/6  )(^x  +  v/y ).     Whence 

(a  —  V'^  )(x/x -f- ^V)  .- 

we  have — — ==  a  —  v'6. 

Vx  +  V^/ 

18.  Divide  a -\- b  —  c  -\-  2^db  by  \/a  +  \/6  —  ^c. 


COMJBINATIONS — INVOLUTION.  201 


OPERATION. 

o  +  2-^06"+ 6  —  c     \-y7i -[-  'Jb  —  j'g 
a  -f-    ^y  ab  —  v'ac        V  a  -\-  's/b  -j-  -v  c 

>/ab  -\-b  -\-  Vac  —  c 

V^a6  -(-  ?)  —  V  oc 

"/ac  -|-  V  6c  —  c 
-s/ac  -f-  ^^^^c  —  c 

19. 

Divide  h^a^-  —  6'^  by  av'(a  +  6)2. 

20. 

Divide  ^a-'  —  ^-  hy  a  —  x.                     Quot,  .    ^  +  '^. 

Na  —  X 

^1 

T^iwl..,      -^-Iby.,     ,^  +  1.           g.r)/./'^'^--'^ 

nJ^  +  1     ^     N/o;-!           ^           ^K'^+1) 

INVOLUTION. 

[Note. — The  principles  heretofore  given  are  sufficient  to  enable  us 
to  involve  radicals,  but  it  may  be  well  to  collect  and  review  them.] 


233,  J* rob,  4.  To  raise  a  radical  to  any  power. 

R  ULE. — Involve  the  coefficient  to  the  required  power, 

AND  ALSO    THE   QUANTITY   UNDER  THE  RADICAL   SIGN,  WRITING   THE 
LATTER  UNDER  THE  GIVEN  SIGN. 

Dem. — This  results  directly  from  the  principles  of  multiphcation  of 
radicals  {230).  Thus,  to  raise  a  \/b  to  the  ?nth  power,  is  to  take  m 
factors  of  a\/b,  which  gives  a^b  X  a-C^b  X  a\/b,  etc.,  to  m  factors. 
But  as  the  order  of  the  factors  is  immaterial  {85)  this  may  be  written 

oxm to  m  factors  X  >7t^X  \^b  X  v^^  to  m  factors.     But  aaa 

to  m  factors  is  by  notation  a™,  and  v''^  X  v^^  X  v'^ to  m  factors 

is  by  {230)  v'^b'".    .-.  The  mth  power  of  a^'b  is  a^'^b'":      q.  e.  d. 

EXAMPLES. 

1.  Raise  |  ^  |  to  the  3rd  power. 


202  CALCULUS    OF   RADICALS. 

MODEL,    SOLUTION. 

Operation.  {ix/Jf  =  iv/f  •  ix/^-Wl  =  ^  •  i  •  i  X  v/f  •  \/|  '  \/l 
=  ^fV^^  or  rizVl  =  ^f  a  v/ia 

Explanation. — The  cube  of  i\/f  is  three  factors  of  i\/|  multiplied 
together;  but  as  the  order  of  the  factors  is  immaterial  this  may  be 
considered  as  3  factors  of  I,  or  -^,  and  3  factors  of  \/|,  or  f  v/f .  Hence 
(3  v^l)^  =  2V  X  5  v^5  =  rl^^v^l-  Which  by  removing  the  denominator 
from  under  the  radical  sign  becomes  ^yjv/lO. 


2.  Raise  2'^8a'-^6  to  the  second  power. 

3.  Raise  -"^^x-y  to  the  5th  power. 

4.  Raise  —  ^v)?  ^^  ^^^  ^^^  power. 

5.  Square  3  —  v/2.  Square,  11  —  6v/2. 

6.  Cube  v^  —  \/3.  Cube,  llv/2  —  Ov's. 


7.  Cube  2^x 


Cube,  S{x  —  y)^^ —  y,oY^{x  —  y)^. 


234,  Cor. — To  raise  a  radical  to  a  power  v)hose  index  is 
the  index  of  the  root,  is  simply  to  drop  the  radical  sign.  Thus, 
the  square  of  v/a6  is  ah,  the  cube  of  "^'Ix/^y  i^lxHj,  the 
square  of  v^ —  2a^6  is  —  1a%  the  5th  power  of  v^a*  —  b-  is 

ScH. — This  process  of  involution  is  a  special  case  under  (194). 


EVOLUTION. 

23tj.  I^rob.  S.   To  extract  any  required  root  of  a  mono- 
mial radical. 

B  ULE. — Extract  the  required  root  of  the  coefficient, 

AND  OF  THE  QUANTITY    UNDER  THE  RADICAL    SIGN    SEPARATELY,  AF- 


COMBINATIONS — EVOLUTION.  203 

FECTING    THE    LATTER    WITH    THE    GIVEN    RADICAL    SIGN.       ReDUCE 
THE  RESULT  TO  ITS  SIMPLEST  FORM. 

Dem. — The  nth.  root  of  aX/b,  signifies  one  of  the  n  equal  factors  of 
a-^b.  Hence  if  we  resolve  a  into  n  equal  factors,  and  v'b ,  into  n  equal 
factors,  the  product  of  one  of  each  is  the  71th  root  of  aZ^b.  Thus  one 
of  the  n  equal  factors  of  a  is   -s/a,    and  of  y^'^b,  v  ^]j^  for  ^  ^b     X 


v'y^  X  ^  -^^ to  71  factors,  is    V  Vb  -  b  -  b  •  bio  n  factors  = 

'l/b{233f  234).     If  now  we  take  the  product  of  one  of  each  set  of 

factors,  that    is    \)\i  V  ^b,  we   have  the  7ith  root  of  aVb,  since  we 
have  one  of  its  n  equal  factors,    q.  e.  d. 
EXAMPLES. 


1.  AVhat  is  the  3rd  root  of  4^  Ba^^? 

MODEL   SOLUTION. 

Opeeation.      V4v/3a3a;  — ^  1^4  V  i^3a-a;  =  t^^  y^''Sa^x  =  ' 

Explanation. —The  cube  root  of  4:^'da'x  is  one  of  the  3  equal  fac- 
tors which  compose  it.     In  order  to  find  this,  I  resolve  each  of  the 

two  factors  4  and  -v  3a'  x  into  3  equal  factors,  and  take  the  product  of 
one  of  each.  4  resolved  into  3  equal  factors  becomes  F  4  X  1^4  X 
1^4.     (A  process  which  in  reality  is  only  indicated.)     In  like  manner 

y/da  'X,  resolved  into  3  equal  factors  becomes  W  V3a^  X  1^^3a^X  V^ct^x 

or  V  f^3a^  X  V  ^Sa^  X  V  fSa^x  since  the  root  of  the  product 
equals  the  product  of  the  roots.     Now  taking  one  factor  out  of  each  of 

these  groups  I  have  F  4  V  ^3a^x,  and  as  three  such  factors  could  be 

formed  from  the  number,  ^  4  \/  ]7  Sa^x  is  the  cube  root  of  4v/3a^x.  But 
this  expression  can  be  reduced  to  a  more  simple  form,  by  observing  first 
that  the  square  root  of  the  cube  root  is  the  6th  root,  and  then  reducing 

fT  to  the  same  radical  form.  Thus  I  have  1^4  V  f  3a^=  fI^y3a-x 
=  ^I^sy-S^=  v'48a^x  by  (233).'' 

*  The  author  can  hardly  refrain  from  apologizing  for  so  elementary  a  demonstra- 
tion at  this  stage  of  the  work.  But  if  the  spirit  of  the  treatment  of  radicals  is  ap- 
prehended, the  reason  for  this  will  be  understood.  If  the  pupil  once  comes  into 
full  possession  of  the  idea  that  factoring  is  the  basis  of  the  whole  subject,  he  has 
th«  key  to  all  its  difficulties. 


204  CALCULUS   OF   RADICALS. 

2.  Extract  the  cube  root  of  va^x^.  Boot,  x^  a\ 

3.  Extract  the  square  root  of  32'^iy2a«.'r^ 

SuG.— The    square    root    of    32l^/iy2a^  =  4n/2  X   v^lS^'  = 
4v^2  X  2a^^3a^=  8a-/ 2-^30^=  8a-^8-5'3^^=8a-^2ia^. 

4.  Extract  the  cube  root  of  -  a^^h. 

8 

5.  Extract  the  4th  root  of  16^2^^^.  Boot,  ^Va^x. 

6.  Extract  the  cube  root  of  (a  +  j^)  Va  +  x. 


Boot,  V  a  ^  X. 

7.  Extract  the  cube  root  of  ovlo-  Boot,  q-y's^. 

8.  Extract  the  square  root  of  7v  Jh-  Boot,  -v/2. 

230,  ScH. — This  operation  is  but  a  special  case  of  qfeding  a  quan- 
tity with  any  given  exponent  {104),  and  the  examples  may  be  performed 
according  to  the  rule  there  given.     Thus,  to  extract  the    cube     root 

of  ^V^ax- ;   putting  it  in  the  form  8  X  (3ax,")^  and  dividing  exponents 

by  3  (multiplying  by  \)  we  have  8-*(3aic")'5  __  2A/3ax-.    The  pupil  may 
solve  the  following  in  this  manner  : 


9.  Extract  the  5th  root  of  ^^1x^^. 
SuG.    -s/32x"^  =(32)*x5.      Multiplying   exponents     by  \   we  have 

(32)"iV  But  (32)'^"' =  1(32)^  =y2.  .• .  V^ ^32^^=  xn/'Z  But 
the  most  simple  way  to  solve  this  particular  case  is\/-v/32x"-  = 
V  ^32xi^=  V2x2  =  x-y%     Or  by  the  .rule  given  {235). 

10.  Extract  the  square  root  of  '^49a2. 

11.  Extract  the  cube  root  of  G4v/8a«. 


COMBINATIONS — EVOLUTION.  205 

12.  Extract  the  5th  root  of  486a  ^ia^  Boot,  3^2^. 


rOR  REVIEW  OR  ADYANCED  COURSE. 

[Note. — This  article  should  not  be  taken  till  after  Quadratic  Equa- 
tions, and  is  not  especially  important  in  an  ordinary  course  at  all.  ] 


237*  Pvoh.  0,  To  extract  the  square  root  of  a  binomial,  one 
or  both  of  luhose  terms  are  radicals  of  the  second  degree. 

Solution. — Such  binomials  have  either  the  form  a  +  n\^b  or 
w-v/a  -4-  ?i'v/6.  Now  observing  that  (x  -j-  y)'^  =  x-  +  2xy  -j-  y^,  we  see 
that  if  we  can  separate  the  first  term  of  any  such  binomial  surd  into 
two  parts  the  square  root  of  the  product  of  which  shall  be  ^  the  other 
term,  these  two  parts  may  be  made  the  first  and  third  terms  of  a  tri- 
nomial (corresponding  to  x-  -f-  2a-?/  -f-  y-),  and  the  middle  term  being 
the  second  term  of  the  given  binomial,  the  square  root  will  be  the  sum 
or  difference  of  the  square  roots  of  the  paxts  into  which  the  first  term 
is  separated. 

EXAMPLES. 

1.  Extract  the  square  root  of  87  —  12^^. 

Solution. — Let  x  =  one  of  the  parts  into  which  87  is  to  be  sepa- 
rated, and  87  —  x  the  other.  Then  we  have  v'a-,(^87  —  x)  =  —  6x^42,  or 
squaring,  87ic  —  x'^  =  1512,  or  ic^  _  87x  =  —  1512.    .  • .  a;  =  63  or  24, 

and  we  have  ^87  —  12^42"  r==  ^63  —  12^42^+  24.  Nom^  ^03"  = 
3  -v/?  and  'v/24  =  2  v^6,  and  as  the  middle  term  of  the  trinomial  is 
negative  and  twice  the  product  of  these  roots,  its  square  root  is  3-^7  — 
2^e.  {123.) 
2.  Extract  the  square  root  of  Sv'q  +  2^12, 
SuG.  3'>/Q  i^  ^^  ^^  separated  into  two  parts.  Let  them  be  x  and 
3-v/6  — X.  Thenx(3-v/6  — x)  =  12.  Whence  x  =  2-^ 6  or  V'g^  and 
v^3v/6  +  2^/12"  r=  ^2Vg  +  2^12"+  v/6  =  ^^2^6  +  ^^6"  = 
t/24 -f  ^Q. 


206  CALCULUS   OF   RADICALS. 

3.  Extract  the  square  root  of  12  —  v^l4U. 

Root,  \/7  —  ^. 
4  Extract  the  square  root  of  11  +  Q^2. 

5.  Extract  the  square  root  of  13  +  2\/30. 


6.  Extract  the  square  root  of  ax  —  ^a^ax  — a-. 

SuG.— Letting  y  be  one  of  the  parts  into  which  ax  is  to  be  separated, 
the  equation  from  which  its  value  is  found  is  y^  —  axy=^  —  a'{ax  —  a-) 
or  —  a^x  -\-  al     Whence  y  =  ax  —  a'^  ox  a"^,  and  the  parts  are  ax  —  a'^ 

and   a-.   Hence  \/ax  —  2a^/ax  —  u^  = 

V{ax  —  a^)  —  '■2a\/ax —  a^  -|-  a'^  =  Vax  —  a-^   —    -v/o^    or   a    — 

7.  Extract  the  square  root  of  2a  +  2^ a-  —  b^. 


IMAGINARY  QUANTITIES. 

238.  An  Imaginary  Quantity  is  an  indicaied 
even  root  of  a  negative  quantity,  or  any  expression,  taken 
as  a  whole,  which  contains  such  a  form  either  as  a  factor 
or  a  term.  Thus  V  —  b,  ^  —  a:\  h^  —  h\  2+^ — 4, 
^ — 6,  3  —  v/ —  1^  etc.,  are  imaginary  quantities. 

230,  ScH.  1. — It  is  a  mistake  to  suppose  that  such  expressions  are  in 
any  proper  sense  more  unreal  than  other  symbols.  The  term  Impos- 
sible Quantities  should  not  be  applied  to  them  :  it  conveys  a  wrong 
impression.  The  limits  of  this  work  prevent  anything  more  than  a 
mere  explanation  of  the  method  of  multiplying  or  dividing  one  imagin- 
ary of  the  second  degree  by  another.  A  fuller  development  of  the 
theory  of  these  interesting  and  important  symbols  will  be  given  in  the 
Higher  Algebra. 

24:0,  ScH.  2.  — A  curious  property  of  these  sjonbols,  and  one  which  for 
some  time  puzzled  mathematicians,  appears  when  we  attempt  to  multi- 
ply -v/  —  a  by  v'  —  a.  Now  the  square  root  of  any  quantity  multi- 
plied by  itself,  should,  by  definition,  be   the  quantity  itseH ;  hence 


COMBINATIONS   OF  IMAGINARIES.  207 

V  —  a  X  '^  —  a  =  —  a.  But  if  we  apply  the  process  of  multiph'- 
ing  the  quantities  under  the  radicals  we  have  's/  ■ —  a  X  ^  —  «  = 
•>/a-  r=  -}-  a,as  well  as  —  a.  What  then  is  the  product  of  x/  —  a  X 
-v/  —  a  ?  Is  it  —  a,  or  is  it  both  -|-  a  and  —  a?  The  true  product  is 
—  a  ;  and  the  explanation  is,  that  \/a~  is,  in  general,  -\-  a  or  —  a.  But 
when  we  know  what  factors  were  multiplied  together  to  produce  a~, 
and  the  nature  of  our  discussion  limits  us  to  these,  the  sign  of  -v/a^  is 
no  longer  ambiguous  :  it  is  the  same  as  was  its  root. 


241.  Projy.  Every  imaginary  term  of  the  second  de- 
gree {and  in  fact  of  every  other  degree)  can  he  reduced  to  the 
form.  mV— 1  in  which  m  is  not  imaginary,  m  may  le  ra- 
tional or  surd. 


Dem. — Let  V  —  X  represent  any  such  expression.  Then  V  —  x 
^^  a/x^  —  1)  =  -J  X'J  —  1,  which  is  the  required  form. 

24:2-  ScH.  3. — By  means  of  this  proposition  and  the  property  no- 
ticed in  Hch.  2,  the  multipUcation  and  division  of  imagiuaries  is  ef- 
fected. 

EX^VMPLES. 

1.  Multiply  4v^^=l5  by  2v/^^. 

Operation.  4^/  —  o  =  4^3  x  V  —  1.  Also  2v/  —  2  =  2-^2 
X  -y"^^.  Hence,  4n/  —  3  X  2^/  —  2!  =  2  X  4x/3  X  -v/^  X 
y  —  1  X  ^  —  1  =  8^  6  X  ^  -  1  X  v/^ITi;  =,  _  8^ ^^ 
since  ^  —  1  X  ^  —  1  =  —  1. 

2.  Multiply  3v/  — 5by4v/  — 3.  Prod,  —  12\/l5. 

3.  Multiply  ^  —  X''  by  v    —  yK  Prod.,  —  xy. 

4.  Multiply  2^^^^  by  3v/'^^  Prod,  —  36. 

5.  Multiply  2v/  — 6,  5v/  — 4,  3^^  —  7  together. 

ProtZ.,  —  60  v^v'  _  1,  or— 60v/—  42. 


208  CALCULUS    OF   RADICALS. 


6.  Multiply  24  +  v/  _  4i)  by  24  —  v^  —  49. 

SuG.— We  have  here  the  product  of  the  sum  and  difference  of  two 
quantities  which  equals  the  difference  of  their  squares.  Hence 
(24  4-  -v/^^l9)  X  (24  —  y^=l9)  =  (24)2  _  (v^_49)2  =  575  _ 
(  _  49)  =  576  -I-  49  =  G25. 

7.  Divide  v/  —  16  by  \/  — 4. 

Opeeation.    ^  —  iG  =  ■i'J  —  1,  and  v/  —  4  =  2^  —  1.    Hence 

■s/  — 16  _  4'v/^ri; 


9 


ScH. — A  superficial  view  of  the  case  might  lead  the  pupil  to  think 
that  the  quotient  was  +  2.     Thus, noticing  that  the  radicals  are  simi- 


-v/— 16  -16  ._ 

lar,  he  might  conclude  that  =  .   I =  V  4  =  +  2,     an 

incorrect  result. 
8.  Divide  Gv^  _  3  by  2v/ —  4.  QwA.,  -^'6. 


9.  Divide  _  v/_  1    by  —  6v/  — 8.  Qiiot,  —  v^S. 


2 

1_ 
18 


10.  Divide  1  +  \/—  1  by  1  —  -^  —  \. 


1  _|-  y  _  1  . 

Opeeation. — Writing  the  example  thus '  and    rational- 

izing the  denominator  by  multiplying  both  terms  of  the  fraction  by 

, 1  4-  2V^:ri  _  1       2x/"^=~l 

1  4-  V  —  1  there  results,  (see  JSc.  6) ..  __      ■., = 3 

The  example  may  also  be  performed  thus  : 

1  _|-  ^^1~I    1^  _  V^iri  -f  1 
1  +  y/  —  1  y  —  1 

Since  1  divided  by  —  »/  —  1  gives  ^  —  1,  as  —  v'  —  1  X  '^ —  1 

=  1.  /■ 


SYNOPSIS. 


209 


Synopsis  for  Eeview. 


Power. — Degi-ee  of.  ]  Sch.  1.  P.  and  E.  correlative. 
Koot. — Degree  of.     f  Sch.  2.  Square,  Cube,  etc. 

[  1     \T  1.  r  Cor.    Transf 'ring 

TT  ^       T  ^      J  o    ti      f-^^'       J    afactorinafrac- 

Exponent  or  Index  ^2.+ Fraction.  ^.^^    ^^^^    ^^^ 

(3. -Int.  or  Fr.        term  to  another. 


Radical 


Real 


j  Rational. 
I  Irrational. 
Imaginary. 

How    indicated  <  ,„ 


Sim.  Rads.     Rationalization.     To  a£fect  with  Expt. 

Involution.     Evolution.     Calculus  of  radicals. 

Frob.  1.  To  any  power.     RuiiE.     Bern.      [Co7\  Signs. 

cl.  +m. 
Prob.  2.  To  affect  with  Expt.  Rule.  Deyn.  4  2.  4-^. 

(3.  —  m  or  -  T* 
Coi\  1.  Terminates. 
Cor.  2.  No.  Terms. 
Cor.  3.  CoeflBcients. 
Cor.  4.  Expt's  in  ea.  term. 
Cor.  5.  Rule. 
Cor.  6.  («— Jf.- 
/Sc^.  Signs. 
Cor.  1.  Same  as  (193). 
Cor.  2.  R.    of  Prod.= 

Prod,  of  Roots. 
Co7\  3.  R.    of    Quot.= 
Quot.  of  Roots. 
(  Sch.  1. 
Prob.  2.  Sqr.  R.  of  Poly.  Rule.  Bern.  ■{  Sch.  2. 

(  Sch.  3. 

Prob.  3.  Sqr.  R.  of  Dec.    No.   Rule.  Beiii. 


Prob.  3.  Binomial  Formula - 


Prob.  1.  Roots  of  Perfect  Pow- 
ers. Rule.  Bern. 


i  Sch.  1. 
I  ^c^.  2. 


Pro5.  4.  CubeR.  of  Poly.    Rule.    Bern.  |  |^^-  2 

(  Sch.  1. 
Pro5.  5.  CubeR.  of  Dec.     No.  Rule.    Bern.  ■{  Sch.  2. 

(  Sch.  3. 
Prob.  6.  Any  R.  whose  index  composed  of  factors,  2  or  3, 
Prob.  7.  Any  Root.  Solution.     [Genl.  Scholium. 


210 


SYNOPSIS. 


SYNOPSIS   FOR   -KEYIWN.— Continued. 

Prob.  1.  Remove  factor.    Rule,  j  Sch.  Simplest  form. 

Dent.  I  Cor.  Denom.  of  surd. 

P7^ob.  2.  Index  composite,  etc.     Rule.     Dem. 

Prob.  3.  To  given  index.     Rule.     Dem. 

^  Prob.  4.  To  common  index.     Rule.     Dem. 

Prob.  5.  To  rationalize  Moii.  Denom.      Rule.     Dem. 

Prob.  6.  To  rationalize  Binomial  Denom.     Rule.    Dem,. 

Prop.  1.  To  rationalize  any  binomial.     Dem. 

Prop.  2.  To  rationalize  any  trinomial.     Dem. 

Prob.  1.   To  add  and  subtract.     Rule.     Dem. 


i  Prop.  1.  Prod,  of  roots=equal  root  of  product.  Dem. 
\  Prop.  2.  Similar  Rads. ,  how  multiplied. 


Rule. 


(  Prob.  2.  To  multiply  Radicals. 

J  Prop.  Quot.  of  roots=root  of  quots, 


Dem. 
Dem. 


q  I  Prob.  3.  To  divide  Radicals 
Prob.  4.  Involution  of  Radicals. 
Prob.  5.  Evolution  of  Radicals. 

VProb.  6.  ^ 


Rule. 
Rule. 
Rule. 


Dem. 

Dem.  [Sch. 
Dem. 


Def. 


'^b  and  '^^a-^'^b,  etc. 

Sch.  1.  Called  Impossible. 
Sch.  2.  Reason  for  name. 


Prob.  1.  To  add  or  subtract.     Rule.     Devi. 
Prob.  2.  To  multiply.     Rule.     Dem. 
Prob.  3.  To  divide.     Rule.     Dem. 


Test   Questions. —By  what  must   numerator   and   denominator  of 
be  multiphed  to  reduce  it  to  a  simple  fraction  ?     Give  the 


^'  .         /-      ^- 

various  significations  of  an  exponent.    Perform  the  operation  x/2  X  1^  3, 

and  explain  the  process.     Repeat  the  Binomial  Formula,  and  by  means 

of  it  expand  (1 — x^)^.     Demonstrate  the  rules  for  square  and  cube 

en         ^^    ^  a  -\-  hx.  —  '>/ (fi  A-h'^x^  ^rri  _i_  M-r,"-  —a 

root.     Show  that — X '  __    n^  a    -f-  t>  .c        a       ^j^^^ 

a  —  hx-{-  y ^2  _j_  ^.^2  5^ 

sign   is   given   to  a  square   root?     Why?     To   a  cube  root?     Why? 
What  is  the  value  of  ic"  ? 

[Note. — Here  ends  the  subject  of  Literal  Arithmetic.  The  student 
is  now  prepared  for  the  study  of  Algebra,  properly  so-called  ;  i  e. , 
The  Science  of  the  Equation.  ] 


PART    II. 

ALGEBRA. 


CHAPTER  I. 

SIMPLE  EQUATIONS. 


SECTION  L 
Equations  with  One  Unknown  Quantity. 

DEFINITIONS. 

1,  Ati  Eqiiatiori^  is  an  expression  in  mathematical 
symbols,  of  equality  between  two  numbers  or  sets  of 
numbers. 

III.  3x  —  2a"y  = — '—  is  an  equation  because  it  is  an  expres- 
sion of  equality  between  3ic  —  2a'^?/  and 

2,  Algehl'Ct  is  that  branch  of  Pure  Mathematics  which 
treats  of  the  nature  and  properties  of  the  Equation  and  of 


*  Do  not  pronounce  this  word  "  Equazion."     For  this  common  error  there  is  no 
authority.     "  Equashun  "  is  the  correct  pronunciation. 


212  SIMPLE    EQUATIONS 

its  use  as   an  instrument  for  conducting  mathematical  in- 
vestigations. 

3,  The  First  If  ember  of  an  equation  is  the  part  on 
the  left  hand  of  the  sign  of  equality.  The  Second 
Jlf ember  is  the  part  on  the  right. 

4,  A  JV^ufnerical  JEquatiou  is  one  in  which  the 
known  quantities  are  r-epresented  by  decimal  numbers  ;  as 
12^2  _  3^  =  48. 

ScH. — This  is  another  instance  of  the  unfortunate,  because  incor- 
rect use  of  the  word  number.  25  is  no  more  a  number,  or  representa- 
tive of  a  number,  than  is  a  or  x.  The  former  is  a  number  expressed 
in  the  decimal  notation,  and  the  latter  in  the  Uteral.  But  perhaps  we 
must  retain  the  term. 

5,  A.  Literal  Equation  is  one  in  which  some  or  all 
of   the  known  quantities  are  represented   by  letters ;    as 

4cj;2  —  2 


ax  ■ —  c  -f  'Sby  = 


8 


G,  The  I>egree  of  an  Equation  is  determined  by 
the  highest  number  of  unknown  factors  occurring  in  any 
term. 

I1.L.  ax  —  tx'  =  c  -|-  £c^  is  of  the  3rd  degree;  a'^x  —  4a;  =  12  is  of 
the  1st  degree  ;  x'^y"-  =  18  is  of  the  4th  degree,  etc. 

7.  A  Simx^le  Equation  is  an  equation  of  the  first 
degree. 

III.     y  ^=:  ax-\-h  is  a   simple  equation,   as  also  is  — f-  4a;  = 

J-.  +  5. 

S,  A  Quadratic  Equation  is  an  equation  of  the 
second  degree. 

9.  A  Cubic  Equation  is  an  equation  of  the  third 
degree.    A  IMquadratic  is  one  of  the  fourth  degree. 


WITH  ONE  UNKNOWN  QUANTITY.  213 

10.    Equations    above  the     second    degree     are    called 
mgher  Equations, 


TRANSFORMATION  OF  EQUATIONS. 

11,  To  Transform  an  Equation  is  to  change  its 
form  without  destroying  the  equality  of  the  members. 

12,  There  are  ybwr  principal  transformations  of  simple 
Equations  containing  one  unknown  quantity,  viz  :  Clearing 
of  Fractions,  Transposition,  Collecting  Terms,  and  Dividing 
by  the  coefficient  of  the  unknown  quantity. 

13,  These  transformations  are  based  upon  the  fol- 
lowing 

AXIOMS. 

Axiom  1.  Any  operation  may  be  performed  iipon  any 
term  or  upon  either  member,  which  does  not  affect  the  value  of 
that  term  or  member,  without  destroying  the  Equation. 

Axiom  2.  If  both  membei^s  of  an  Equation  are  increased 
or  diminished  alike,  the  equality  is  not  destroyed. 


11,  JProh,    To  Clear  an  Equation  <rf  Fractions. 

rule. muxittply  both  membees  by  the  least  oe  lowest 

common  multiple  of  all  the  denomenatoes. 

Dem. — This  process  clears  the  Equation  of  fractions,  since,  in  the 
process  of  multiphing  any  particular  fractional  term,  its  denominator 
is  one  of  the  factors  of  the  L.  C.  M.  by  which  we  are  multiphing;  hence 
dropping  the  denominator  multiphes  by  this  factor,  and  then  this 
product  (the  numerator)  is  multiphed  by  the  other  factor  of  the 
L.  C.  M. 

This  process  does  not  destroy  the  Equation,  since  both  members  are 
increased  or  diminished  alike. 

EX^VMPLES. 

rp  rv*  y*  2 '7*        1  Q 

.  1.  Clear  the  equation  9  +  0+^  = — "o —  of  fractions. 


214  SIMPLE   EQUATIONS 

Model  Solution.  6  is  the  L.  C.  M.  of  2,  3,  6,  and  3,  the  denomina- 
tors. Now,  it  is  evident  that,  if  the  first  member  of  this  equation  is 
equal  to  the  second,  6  times  the  first  member  is  equal  to  6  times  the 
second  ;  hence  I  can  multiply  both  members  by  6  and  not  destroy  the 

X 

equahty.     In  order  to  multiply  -^  hj  6  1  drop  the  denominator,  thus 

multiplying  the  term  by  2,  and  then  multiply  x  by  the  other  factor  of  6, 

obtaining  3x.     In  like  manner  dropping  the  denominator  of  '—  and 

o 

multiplying  the  product,  x,  by  2,  I  multiply  the  term  —  by  6,  and  have 

o 

2x.     Dropping  the  denominator  of  —  multiplies  this  term  by  6.  Hence 

D 

6  times  the  first  member  of  the  given  equation  is  3x  -}-  2x  -{-  x.  Six 
times  the  second  member  is  found  by  dropping  the  denominator,  3, 
and  multiplying  by  2,  giving  4x  -\-  6.  Hence  3x  -{- 2x -{-  x  =  4:X  -{-  Q  ; 
and  the  denominators  have  all  been  dropped  without  destroying  the 
equality,  as  both  members  of  the  equation  have  been  increased  alike. 

III. — An  equation  is  apt- 
ly compared  to  a  pair  of 
scales  with  equal  arms, 
balanced  by  weights  in  the 
two  pans. 

Now,  if  the  weights  in 
the  scale  pans  balance  each 
other,  that  is,  are  equal,  and 
we  multiply  the  weights  in 
each  pan  by  6  (or  any  other  number),  the  balance  (equahty ")  will  still 
be  preserved.  Or,  if  we  increase  or  decrease  the  weights  in  both  pans 
equally,  the  balance  (equality)  will  not  be  destroyed. 

2.  Clear  —  _  —  -f  1  =  —of  fractions.  • 

nesuU,  8x  —  15^7  +  12  =  GQ. 

'  3.  Clear  10  +  ?^-=^  =.  i-±-^  -  3^  of  fractions. 
5  2 

Eesult,  100  +  4^  —  10  =  5  +  5^  —  30^. 

.     ^,          X  —  1        X  —  2                3.77  —  1       4   —  .r     , 
4.    Clear    — f-  -^  =  — 7; f- 


2  3         '  6^3 

fractions. 


WITH   ONE   UNKNOWN   QUANTITY.  'Mb 

X 2 

SuG. — The  multiplier  is  6.     In  multiplying  the  second  term,  — -— , 

o 

it  should  be  borne  in  mind  that  the  —  sign  preceding  this  compound 
term,  shows  that  the  term  as  a  whole  is  to  be  subtracted.  Hence  when 
this  term  is  -^Tritten  -udthout  any  mark  of  aggregation,  its  signs  are  to 
be  changed,  as  m  removing  terms  from  a  bracket  preceded  by  a  minus 
sign.  The  equation  cleared  of  fractions  is  ox  —  3  —  2x  -j-  4  -}-  6x  =: 
Sx  — 14-8  —  2x.     Wliy  are  not  the  signs  changed  in  the  last  term, 

4 a-  X 2 

■ — ; —  ?      Are  all  the  signs  changed  in  the  term  '—— —  ?  Yes.  WTiat  be- 

comes  of  the  —  sign  before  this  fraction  in  the  given  example  ?  It  is 
dropped  after  the  operations  signified  by  it  have  been  expressed  in  de- 
tail. We  might  wTite  the  equation  cleared  effractions  thus:  3x —  3  — 
(2x  —  4)  +  6ic  =  3x  —  1  -f  8  —  2x,  the  term  2x  —  4  being  still  taken 
in  the  aggregate.  Now  removing  the  parenthesis  (give  the  reason)  we 
have  3x  —  3  —  2x  -{-  4  -f-  6x  =  3x  —  1+8  —  2x,  as  above. 

Neglect  to  make  this  change  of  signs  is  one  op  the  most  com- 
mon  MISTAKES   or  BEGINNEES. 

5.  Clear 1 —  =  4c '■ of  fractions. 

m        am        m^  a-m 

Suggestion. — The  multiplier  is  a-m'^. 

Result,  a-mx  -f  amx  —  a-x  =  Aa-cm^  —  Smx  +  mn. 

6.  Clear    y  —  — r =  -  of  fractions. 

4  2  6 

BesuU,  dx  —  6x  +  18  =  2a. 

2^       ^ H^        3 

-  7.    Clear —— \ :=  1  —  x  oi  fractions. 

3a  zab         a- 

BesuU,  Aabx  —  dax  +  3a  +  18b  =  Ga^ft  —  6a'^bx. 

^     ^1               .r                           8             a  —  ^^         w     /.   p 
8.    Clear    z x  -\ = —  1  oi  irao- 

a  —  0  a   -{-    b        a+6 

tions. 
Suggestion. — The  multipher  is  a-  —  62. 

BesuU,  ax  +  bx  —  a^x  +  b^x  -\-  3a  —  36  = 

(a  —  b)-^  —  a-^  +  b'^ 

QuEKT. — Why  are  the  4th  and  last  terms  -f  ? 


216  SIMPLE   EQUATIONS 

9.    Clear = —  of  fractions. 

jc  ■ — ■  c        X  +  "Zc 

Result,  ax  +  2ac  —  hx  —  Ihc  =  ax  -\-  bx  —  ac  —  be. 

^T  X 

10.    Clear    —-7  =  3  +  - — '—r  of  fractions. 

a  —  2b  la  —  b 

Result,  Aax  —  2bx  =  6a''  —  15ab  +   6t'  +  ax  —  2bx, 


TRANSPOSITION. 
IS,  Transposing  a  term  is  changing  it  from  one 
member  of  the  equation  to  the  other  without  destroying 
the  equality  of  the  members. 


16,  JProb,   To  irani^pose  a  term. 

RULE. — Drop  it  from  the    member  in  which    it   stands 

AND    INSERT   IT   IN   THE    OTHER    MEMBER   WITH  THE  SIGN  CHANGED. 

Dem.— If  the  term  to  be  transposed  is  -}-,  dropping  it  from  one 
member  diminishes  that  member  by  the  amount  of  the  term,  and 
writing  it  with  the  —  sign  in  the  other  member,  takes  its  amount  from 
that  member ;  hence  both  members  are  diminished  ahke,  and  the 
equality  is  not  destroyed.     (Kepeat  Axiom  2. ) 

2nd. — If  the  term  to  be  transposed  is  — ,  dropping  it  increases  the 
member  from  which  it  is  dropped,  and  writing  it  in  the  other  member 
with  the  -\-  sign  increases  that  member  by  the  same  amount ;  and 
hence  the  equality  is  preserved.     (Repeat  Axiom  2. ) 

EXAMPLES. 

1.  Given  the  equation  3  +  2jt  —  5  =  12  —  ^x  to  trans- 
pose so  that  all  the  terms  containing  the  unknown 
quantity,  x,  shall  stand  in  the  first  member  and  the 
known  terms  in  the  second  member. 

MODEL    SOLUTION, 

OPEIIA.TION.  3  -f-  2x  —  5  =  12  —  Ax 

2x  +  4ic  ==  12  —  3  +  5. 

Explanation. — Dropping  3  from  the  first  member  diminishes  that 
tnember  by  3 ;  hence  to  preserve  the  equality,  I  subtract  3  from  the 


V      WITH  ONE  UNKNOWN  QUANTITY.  217 

second  member,  or  indicate  the  subtraction  by  writing  it  in  the  second 
member  with  the  —  sign.     Thus  the  term  3  is  transposed. 

Dropping  —  5  from  the  first  member  increases  that  member  by  5  ; 
and  hence  to  preserve  the  equality  I  add  5  to  the  second  member. 
Thus  the  term  —  5  is  transposed. 

Dropping  —  4ic  from  the  second  member  increases  that  member  by 
4x,  hence  I  increase  the  first  member  by  adding  4x  to  it,  and  thus  pre- 
serve the  equality. 

I  have  thus  arranged  the  terms  so  that  all  those  containing  the  un- 
knovni  quantity  stand  in  the  first  member,  and  all  known  terms  in  the 
second  member  ;  and  yet  I  have  preserved  the  equality.  This  is  called 
transposition. 

III. — This  operation  can  be  illustrated  by  the  scales  on  page  214. 
Suppose  the  positive  terms  to  represent  weights  and  the  negative  terms 
some  forces  lifting  on  the  scale-pans.  [This  it  wiU  be  remembered  is 
the  significance  of  the  signs  -\-  and  —  as  signs  of  character.  Thej^  re- 
present quantities  opposed  in  effects.  ]  Taking  the  3  from  the  first 
member  corresponds  to  taking  off  so  much  weight.  This  can  be  com- 
pensated, so  as  to  keep  the  scales  in  equihbrium  by  applying  a  Ufting 
power  of  3  to  the  other  side,  which  is  symbolized  by  —  3  on  that  side. 
Again  taking  —  5  from  the  first  member  is  like  taking  away  a  lifting 
power  of  5,  which  can  be  compensated  by  putting  a  weight  of  5  on  the 
other  side  (-f-  5).  In  like  manner  the  transposition  of  any  term  can  be 
illustrated.  In  fact  all  operations  upon  equations  can  he  illustrated  i7i  a 
similar  way. 

In  the  following  examples  transpose  the  unknown  terms  to  the  first 
member  and  the  known  to  the  second. 

2 .  Given  5x  —  12a  +  3c  —  2x  =  4:X  —  2x  +  4:a  to  trans- 
pose as  above. 

Besult,  5x  —  2x  —  4.x  +  2x  =  Aa  -{-  12a  —  3c. 

3.  Given  100  +  4^  —  6  =  5^7  +  5  —  30a:  to  transpose  the 
terms  containing  x  to  the  first  member  and  the  others 
to  the  second. 

Besult,  4.x  —  5x-i-S0x  =  5  —  100  +  6. 

15 

4.  Transpose  as  above  __- 

A 

Besult,  ~-roc- 
5 


3.r 

7 

1 

11 

"2 

+  10- 

~8' 

~2' 

"T 

x-{-x. 

3j7 

1 

7 

15 

■  X- 

~"T" 

=  2  + 

8~ 

-10- 

■  2" 

218  SIMPLE   EQUATIONS 

5.  Transpose  as  above  3a6 — 4:ax  4- ISbx"^  =  — '- x^ 

c  5 

2(7.77  2ft^ 

Result,  186a;2  +  ^2  —  4^^ _'  :== 3,3,^, 

5  c 


SOLUTION  OF  SIMPLE  EQUATIONS  WITH  ONE  UNKNOWN 
QUANTITY. 

17*  To  Solve  an  Equation  is  to  find  the  value  of  the 
unknown  quantity  :  that  is,  to  find  what  value  it  must  have 
in  order  that  the  equation  be  true. 

III. — In  the  equation  Ax  —  2  =  2aj  +  4,  if  we  caU  x,  3,  the  first 
member  is  10,  and  as  the  second  is  also  10  for  this  value  of  x,  the 
equation  is  true.  But  if  we  try  any  (or  every)  other  number  than  3 
for  X,  we  shall  find  that  the  equation  will  not  be  true.  Thus  trying  4 
for  X,  we  find  the  first  member  14  and  the  second  12  ;  and  the  equation 
is  not  true.  Again,  try  5.  The  first  member  becomes  18  and  the 
second  14,  and  the  equation  is  not  true. 

Let  the  student  see  if  he  can  ascertain  by  inspection  what  are  the 
values  of  x  in  the  following  : 

X  +  3  =  3ic  +  1, 
2a:  =  30  —  x. 

Though  these  equations  are  very  simple,  it  is  probable  that  it  will 
take  the  student  some  time  to  guess  out  the  values  of  x  which  make 
them  true.  Thus,  trying  the  first,  2  makes  the  first  member  5  and  the 
second  7  ;  and  the  equation  is  not  true  for  this  value.  If  he  tries  3 
the  result  is  worse  than  before.  But  1  makes  each  member  4,  and  for 
this  value  of  x  the  equation  is  true,  and  for  no  other. 

But,  if  it  is  so  difficult  to  hit  upon  just  the  value  of  x  which  is  re- 
quired to  make  so  simple  an  equation  true,  the  task  would  be  quite 
hopeless  in  such  an  one  as 

3x  — 1        13  — a;__    7x        ll(a;-j-3) 

5  2  T  6 

Yet  we  have  a  very  simple  method  of  solving  any  such  equation  so 

as  to  tell  certainly  and  easily  what  the  value  of  x  is.     This  process  is 

now  to  be  explained,  and  is  called  Solving  the  Equation,  or  sometimes, 

thft  Besolution  of  the  Equaiion. 


WITH   ONE   UNKNOWN  QUANTITY.  219 

18,  Au  equation  is  said  to  be  Satisfied  for  a  value  of 
the  unknown  quantity  which  makes  it  a  tru©  equation  :  i.  e., 
which  makes  its  members  equal. 

10.  To  Yeriffj  an  equation  is  to  substitute  the  sup- 
posed value  of  the  unknown  quantity  and  thus  see  if  it  sat- 
isfies the  equation. 


20,  JProb,  1,  To  solve  a  simple  equation  with  one  un- 
known quantity. 

RULE  1. — If  the   equation  contains  fkactions,   ciiEAR  rr 

OF  THEM   BY   Art   Itt. 

2.  Teanspose  all  the  teems  involving  the  unknown 
quantity  to  the  fiest  membee,    and  the   known  teems  to 

THE    SECOND    MEMBEE    BY    Art.    15. 

3.  Unite  ajul  the  teems  containing  the  unkno^^tn  quan- 
tity INTO  one  by  addition,  AND  PUT  THE  SECOND  MEMBEE  IN- 
TO its   SOIPEEST  FOEM. 

4.  Dr^^IDE  BOTH  MEMBEES  by  the  COEFFICIENT  OF  THE  UN- 
KNO^^^   QUANTITY. 

Dem. — The  first  step,  clearing  of  fractions,  does  not  destroy  the 
equation,  since  both  members  are  multiphed  by  the  same  quantity 
(Axiom  2). 

The  second  step  does  not  destroy  the  equation,  since  it  is  adding  the 
same  quantity  to  both  members,  or  subtracting  the  same  quantity  from 
both  members  (Axiom  2). 

The  third  step  does  not  destroy  the  equation,  since  it  does  not  change 
the  value  of  the  members  (Axiom  1). 

The  fourth  step  does  not  destroy  the  equation,  since  it  is  dividing 
both  members  by  the  same  quantity,  and  thus  changes  the  members 
ahke  (Axiom  2). 

Hence,  after  these  several  processes,  ve  still  have  a  true  equation. 
But  now  the  first  member  is  simply  the  unkno-wn  quantity,  and  the 
second  member  is  all  known.  Thus  we  have  what  the  unknown  qiian- 
tihj  is  equal  to  ;  i.  e.,  its  value. 


220  SIMPLE  EQUATIONS 

21,  ScH.  1. — It  mast  be  fixed  in  the  pupiVs  mind  that  he  can  make  but 
two  classes  of  changfs  upon  an  equation :  viz.,  Such  as  do  not  aitect 

THE  VALUE  OF  THE  MEMBEES,  Or  SUCH  AS  AEEECT  BOTH  MEMBEBS 

EQUALLY.     Every  operation  must  be  seen  to  conform  to  these  conditions. 

EXAMPLES. 

1.    Solve    — ^ 1 \-z  =  — :: — .and  verify  the 

Z  O        O         O 

result. 

MODEL  SOLUTION. 

cc  +  1    ,    3cc  — 4   ,1        6a; -f  7 

OPERATION.  (1)   -   -    • 1 ^ '     8   ~  8 ' 

(2) 20d;  -I-  20  4-  24a;  —  32  4-  5  =  30x  +  35, 

(3) 20a;  +24a;  —  30x  =  35  —  20  +  32  —  5, 

(4) 14x  =  42, 

(5) a;  =  3. 

Explanation. — I  first  clear  equation  (1)  of  fractions  by  multiplying 

both  members  by  the  L.  C.  M.  of  its  denominators,  which  is  40.     This 

x  +  1 
does  not  destroy   the  equation  (Axiom  2).     I  multiply  — ~ —  by  40  by 

dropping  its  denominator  2,  thus  getting  a;  +  1,  as  the  result  of  multi- 
plying by  2,  and  then  multiply  a;  -j-  1  by  20,  getting  20a;  -{-  20.  [In 
like  manner  explain  the  entire  process  of  clearing  of  fractions.  ] 

Having  cleared  the  equation  of  fractions  I  have  (2).  I  now  transpose 
the  terms  containing  x  to  the  first  member  and  the  known  terms  to 
the  second  member.  Thus,  dropping  30x  from  the  second  member 
diminishes  it  by  that  amount,  whence  to  preserve  the  equahty  of  the 
members  I  subtract  30x  from  the  first  member,  i.  e.,  write  it  in  that 
member  with  its  sign  changed.  [In  like  manner  explain  the  transpo- 
sition of  each  term.  ] 

I  know  equation  (3)  to  be  true,  since  I  have  changed  both  members 
alike,  that  is  have  added  to  and  subtracted  from  both  members  the 
same  quantities.  I  now  add  together  the  terms  of  the  first  member, 
which  does  not  affect  the  value  of  the  member,  and  have  14x.  In  the  same 
manner  uniting  the  terms  of  the  second  member  does  not  alter  its  value; 
hence  14x  =  42.  Finally,  I  divide  both  members  by  14  and  have  x  =  3. 
This  operation  does  not  destroy  the  equation,  since  both  members  of 
the  equality  14x  =  42  are  divided  by  the  same  number  (Axiom  2). 
Henoe  3  is  the  value  of  a;. 


WITH  ONE   UNKNOWN  QUANTITY.        221 


VEELFICATION. 

I  will  now  see  by  actual  trial  if  3  does  satisfy  the  given  equation. 
Substituting  3  for  x,  I  have 

3  +  1,    3x3-4,1        6x3+7 
— ! U  -  = ■ — ,  or 

4,5,1        25 

2  +  ^+8-^-r'"^ 

2  +  1  +  J  =  3^,  or 

o 

3  +  -=  3s,   the  members  of  which  are 

o 

identical;  and  the  value  of  x  is  verified. 

ScH.  2. — The  pupil  must  not  understand  that  the  verification  is  at 
all  necessary  to  prove  that  the  value  found  is  the  correct  one.  This  is 
demonstrated  as  we  go  along,  in  obtaining  it.  The  object  of  the  veri- 
fication is  to  give  the  pupil  a  clearer  idea  of  the  meaning  of  an  equa- 
tion, and  to  detect  errors  in  the  work. 

2.  Solve  and  verify  — =  — — - .  Result,  x  =  l. 

•^2  8 

2-4-1       7-4-5       3       12 
Veeification.     — +—  =  — - —  or  -  =  ■^,     Whence  it  appears  that 

2  b  2         o 

1  satisfies  the  equation. 

\  3j;    2  1     -\-    X 

3.  Solve  and  verify  1x  —  -  =  —^ 1 — . 

Result,  X  =  2. 

.    ^.        2(^  —  1)       9       J- +  3         4 

4.  Given r —  -  =  — = to  lind  x. 

a  5  7  5 

Result,  X  =  4:. 

5.  Given  Tjt  +  6  —  3^  =  56  +  2.r  to  find  x,  and  verify. 

6.  Given \-  6i/  =  ■ — ■   to  find  y. 

n    r.'        0-.    ,    3-^  —  11       5.r  — 5   ,  07  —  7.77^     .    . 

7-  Given  21  H — —   = \ —  to  find  x. 

lb  o  2 


Result,  X  =  9. 


8.  Given — — ^+  Gx  =  ^ to  find  x. 

20  5 


222  SIMPLE  EQUATIONS 

9.  Given  25(ila;  —  8)  =  18(12.7;  +  2)  to  find  x. 

10.  Given  ?^  +  2^  +  H  =  ?  +  17  to  find  x. 
5  4 


11.  Given  y  -\ — -  =  — - — -  to  find  y 

o  A 


Result,  X  =  10. 

y- 

Result,  y  =  3f . 


.,   13  —  31-      17  —  3^ 
Veeotcation.     3t  H r. = :^ .  or 

o.    .   9f         13^ 
3t  4-  3^  =  2-.  or 

15i  +  ^  =  6f,or 

—I  =  6f ,  whicli  is  a  true  equation,  since 
20^  -^  3  is  ^. 

In  verifying  it  is  not  well  to  go  through  the  processes  of  clearing 
of  fractions,  transposition,  etc.,  but  rather  keep  the  terms  as  distinct 
as  possible,  and  reduce  each  member  separately  to  a  form  so  simple 
that  they  can  be  seen  to  be  equal. 

z 2  5  +  z 

12.  Given — [-32;=  — \-  1,  to  resolve  and  verify  as 

above.  Result,  z  =  ly^. 

SuG. — Id  such  cases  it  will  be  found  more  simple  to  use  the  value  of 
the  unknown  quantity  as  an  improper  fraction  in  verifying.     Thus  in 

2  5. 2       75        5  4_  3^4 

the  last  we  have  ~ 1 =  — ~\  '      -I-   1,  which    reduces    to 

3         '  17  2  ' 

^_3_      75__55^      17_        72__72^ 
17  "^  17  ~  17  "^  17  '  °^  17  ""  17  ■ 

13.  Given  3y  +  ^-^  —  3  -=  3  — 10//_  ^^  ^^  resolve  and 

21 

verify  as  above.  Result,  y  =  ——. 


14.  Given 


106 

3(;r  — 1)      2(^  — 2)_2(2  — x) 
5         "^        15        ~         3        * 

Result,  X  =  — -. 


WITH  ONE  UNKNOWN  QUANTITY.         223 

2x  +    6 

15.    Find    the    value    of     x    vn.    Zx   -\- =  5  + 

5 

llj:  —  37 


'A 

SuG. — Having  cleared  this  equation  of  fractions,  transposed  the  un- 
known terms  to  the  first  member,  and  the  known  to  the  second,  there 
results  —  21a;  =  —  147.  Hence  to  obtain  a  positive  result  divide  both 
members  by  —  21,  and  x  =  7. 

22,  Cor.  1. — All  the  sigits  cf  the  terms  of  both  members  of 
an  equation  can  be  changed  from  +  to  — ,  or  vice  versa,  with- 
out destroying  the  equality,  since  this  is  equivalent  to  multiply- 
ing or  dividing  by  —  1. 

16.  Find  the  value  of  a;  in  ^  -| ~ =  — 2. 

o  o 

Result,  X  =  4:. 

17.  Given  'IszJji  +  ^Jl^  =  21  +  ?^^  to 

2  o  lb 

find  y.  Result,  y  =  9. 

5   I    2  z 2 

18.  Find  the  value  of  z  in  1  H ^r—  =  3z  -\ —  (See 

25 

Ex.  12.)  Result,  z  = —- 

19.  Find  the  value  of  a;  in  ^-^  +  ^-i^  =  16  —  ^-. 

2  3  4 

J.  I  3 
SuG. — In  clearing  of  fractions  be  careful  to  notice  the  term j — 

As  this  term  is  to  be  subtracted,  when  the  sign  of  aggregation  (the  hne 
between  the  terms  of  the  fraction)  is  dropped,  the  signs  of  the  separate 
terms  must  be  changed.  The  equation  when  cleared  is  6x  -f-  6  4"  ^^  7!" 
8  =  192  —  3x  —  9  ;  and  x  =  13.     It  is  well  for  the  pupil  to  explain 

X  4-  3 
such  cases  thus  :  Having  multiplied  — - —  by  the  L.  C.  D,  12,  I  have 

3x  -f  9  ;  but  this  is  to  be  subtracted,  hence  I  annex  it  with  it  signs 
changed,  as  subtraction  is  performed  by .  When  such  oppor- 
tunities present,  the  teacher  should  frequently  require  the  demonstration 
of  the  rule  or  principle  involved. 


224  SIMPLE    EQUATIONS 

20.  What  value  for  z  satisfies  z  +  =  12  ~  ? 
[Note. — Verify  the  answer  in  this  and  the  following  six  examples.] 

21.  Find   the   vahie    of    z    which   satisfies    the   equation 


4  3  12 

22.  Given  !^  -  ?ili:^  =  ^^  _  2|,  to  find  .. 

14  iil  4  ^ 

23.  Given  -|-(7a;  +  9)  +  tV(249  —  9a:)  =  l{9x  —  13)  + 
i(3x  +  -l),  to  find  a:. 

24.  Given  .1(0;  +  1)  +  i(^  +  2)  =  16  —  ^(^  +  3),  to 
find  X. 

25.  Given  l{3x  —  3)  —  U^x  —  4)  =  5^  —  ^27  +  4.x), 
to  find  or. 

26.  Given  1(8  —  a;)  +  ^  —  If  =  i(^  +  6)  —  ^x,  to 
find  07. 

27.  Given 5x  =  6,  to  find  x. 

a 

Operation. — Multiplying  by  a  we  have  3a-\-x  —  5ax  =  6a.  Trans- 
posing, X  —  5ax  =  3a,  or  (1  —  5a)x  =  3a.      Dividing  by  1  —  5a,  x  = 

3a 
I  —   5a" 

[Note. — It  is  the  universal  experience  of  pupils  that  they  continue 
to  find  difficulty  with  Literal  Equations  even  after  they  are  quite  fami- 
liar with  Numerical  ones,  such  as  the  26  above  given.  This  difficulty 
MUST  he  overcome.  No  one  has  caught  the  true  spirit  of  mathematical 
reasoning  till  the  literal  notation  is  seen  and  felt  to  be  more  simple 
than  the  decimal.] 

28.  What  value  for  x  satisfies  ax  -\-  b  =  ex  -^  d?    Verify  it. 

d  —  b 
Ans.,  X  = 


WITH   ONE   UNKNOWN   QUANTITY.  225 

Vekitication. — Substituting  the  value  of  x  for  x  we  have  a 4- 

a—  c 

b  =  c 1-  d.      Performing    the    multiplications    indicated,    and 

a —  c 

reducing    each    member    to    an    improper    fraction,    this    becomes, 

ad  —  ab  +  ab  —  be       cd  —  be  4- ad  —  f'cZ     ._  ,        ,    ,      ,    , 

■ = •    Now  —  ab  and  4-  ab  de- 

a  —  c  a  —  c 

stroy  each  other  in  the  first  member,  and  -|-  cd  and  —  cd  in  the  sec- 
ond.    Hence  we  have  = ,    which  is  evidently  true. 

a  —  c  a  —  c 

X      1 

29.  What  value  for  x  satisfies  ax  -{-  b  =  -  -{-  -'^ 

a      0 

ail  —  b^) 

30.  Solve    ^^  +  ^^=^-f^^^^  =  "~^+'  +  ^'^- 

oca  aJjc 

a^c  +  ab-  -f  bc'^  —  a  —  b  —  c 


Result^  X 


ac  +  a6  -f  6c  —  1 


oi     -ci-   J       £                   a^  —  ^bx                              ebx  —  Ba-' 
31.    Find  X  from  ax ab^  =  bx  -\ ■ 

a  ^        2a 

bx  +  4a 

4«7)2  —  1  Oa 


Besult,  X 


4a  — 36 


32.  Solve    = '-  and  verify  the  result. 

in  c 

33.  Given  aJ7  +  m  =  6a;  +  ?i  to  find  07.     Result,  x- 


a  —   6 

34.  Given  60;  +  2.r  —  a  =  3j7  —  2c  to  find  x. 

Result,  X  = '-. 

6—1 

35.  Given  a^  +  62  =  6.r  +  a^  to  find  x. 

Result,  X  =  a  -{-  b. 

36.  Given  -  +  -  =  c  to  find  x.         Result,  x  = -. 

ax  ,  ac  —  1 


226  SIMPLE    EQUATIONS 

37.  Given  — - — —  =  10a  +  116  to 

6  D  "^ 

find  X.  Result,  x  =  25a  +  246. 

38.  Given he  =  m  4- -  io  find  x. 

X  X 


39.    Given  -  —  1  +  3a6  =  —  to  find  a;. 
a  c 

Result,  X 


ah  —  1 

Result,  X  = 

be-\-  m 

ac(l  —  Sab) 


c  —  ad 


40.    What   is  the   value    of    x  in  — 1- 


6^  +  3 

257  +  4 


23*  ScH.  3. — ^It  is  not  always  expedient  to  perform  the  several  trans- 
formations in  the  same  order  as  given  in  the  rule.  The  order  there 
given,  and  illustrated  in  the  preceding  examples,  will  always  effect  the 
solution,  but  the  process  may  often  be  much  shortened  by  the  exer- 
cise of  a  little  ingenuity.  The  pupil  should  bear  in  mind  that  the  ul- 
timate object  is  to  so  transform  the  equation  that  the  unknown  quantity 
will  stand  alone  in  the  first  member,  taking  care  that,  in  doing  it, 
nothing  is  done  which  will  destroy  the  equality  of  the  members. 

An  expeditious  method  of  solving  the  last  example  is  as  follows  : 

2\x 39 

Multiplying  by  9,  we  have  Q>x-\-l ^     '  =  6x  -j-  12.       (The  term 

7;,.    ]^3 

~ — 7——  is  multiphed  by  3  by  dividing  the  denominator,  and  by  the 
6x  -f-  3 

other  factor,  3,  of  9,  by  multip>lying  the  numerator.)     Now  dropping 

21a*, 39 

6cc  from  both  members  and    subtracting  7,   we  get   — ^ — -— ~  =  5. 

Ax  — j—  1 

Whence  x  =  4. 

,n     n-         4.r+3       7.r  — 29        8^  +  19,      .     , 
41.  Given  -^-  +  ^^^—3-  =  -^  to  find  x. 

4x4-3 
SuG. — Multiply  numerator  and  denominator  of  the  term  — - —  by  2, 


WITH  ONE  UNKNOWN  QUANTITY.        227 

transpose  and  unite  it  to  the  second    member,  and  there    results 

Ix  —  29       13       ^^^^ 

—  =  -— .     Whence  x  =  6. 

5x  —  12       18 

..     _.        9.r:  +  5  ,  8a;— 7       36.^;  +  15  ,10^,       .    , 

42.  Gwen-^  +  g;^^=-^^— +  — ,  to  find  x, 

and  verify  the  result. 

.or.-        6^  +  7      2^  —  2       2^  +  1     •.,  ,        .. 

43.  Given  — -— =  — - — ,  to  find  x,  and  verify 

15  ix  —  b  6 

the  value. 

..    ^.         6a: +  1       2^  —  4       2^  — 1  .     _    ,  ... 

44.  Given  — -— — ,  = — ,  to  nnd  x,  and  verity 

15  Ix — 16  5 


the  result.  Result,  x  =  —  2. 

2-f-l       —4  —  4  _— 4  — 1       _ii_i 
15       ~_14  — 16~       5       '  ^^      15        15 


—  12-4-1        —4  —  4        —4  —  1            11         4 
Verification.     — -! — —z  = = ,  or  —  -— 


45.  Given  —^ — '- — -H-  H ;  =  8,  to  find  x  and  verify  the 

X  +  3         ^+1  ^ 

value. 

XX  Qi 

46.  Solve  for  x  the  equation  -  + =  •; . 

a      0  —  a       0  -\-a 

SuG. — Clearing  this  of  fractions  and  solving  in  the  ordinary  way,  we 
have  h'-x  —  a'^x  -j-  ohx  -\-  a'^x = a-b  —  a^,  or  {b^-]-ab)x = a"b  —  a\     Whence 

a-b  —  a^        a-(b —  a) 

g.  —^ or  — — ■ . 

5^  _|-  ab'        b{b-\-a)' 
A  more  elegant  solution  is  obtained  by  first  uniting  the  two  terms 

«     ,           ,           ,  .  ,         ,       ,■■              ,.       bx  —  ax  A-  ax         a 
of  the  first  member,  which  makes  the  equation ; =-— — , 

a{b  —  a)  b-\-  a 

a-{b  —  a) 


Whence  x  = 


a(b  —  a)         b  -\-  a  b{b  -j-  a) 

47.  Solve  -{x  —  -)  _  -(^  _  -)  +  -  (^  _  -)  ==  0. 

SuG. — Performing  the  multiplications    and    transposing,   we  have 
X        x   ,   X        a         a         a  5  2 

2-^  +  4=  6--12+^'^'l2^  =  -15"-      ^^"^""  ^^'^^  ^y 

r  Q 

j-j-  we  have  x  =  --r=a.     The  pupil  should  solve  this  in  the  ordinary 
way  and  compare  the  processes. 

A" 


228  SIMPLE   EQUATIONS 

48.  Solve  ^  -  ^-  =  *—  ^.  .    Result,  y  =  % 

0        a      a         c  ad 

49.  Solve  ^^  =  3+      ^ 


a  —  26  2a  —  6 

2b^ 

Result.  X  =  2a  —  5b  -\ . 

a 

X 

SuG. — Transpose r  to  the  first  member  and  unite  the  terms, 

2a  —  6 

giving  -— X  =  3.     .  • .  ic  =  2a  —  56  +  — -. 

2a2  —  dab  -{-  26^  a 

50.  Solve  3^  —  a  =  X — . 


^(a  — 6)                a6       j: 
51.  Solve  -- — — -  =  a  + . 

ax  —  hx  ,  X           ,   ah           3a  —  36  4-  2  4a  +  a6 

Sue.     _^_  +  -  =  <,+  -,or x=-^-. 

ia+ab  6  _    3a(4  -f  6) 

»c  =^ X 


52.  Solve 


4  3a— 36-f-2       6^a— 6)4-4 

3.V  —  a       y+  2b       ly      a 


c        4 

8&2  —  4ac  +  abc 
Kesult,  y  - 


12(26  —  c) 

53.   Solve  \{x  —  a)  _l(2a;  — 36)  — l(a  — .r)  =6  — fa. 
o  4  M 

Result,  X  =  j6. 


3     ' 

that  ^  =  0. 

55.  Solve  -^— - 

bx 

,  ax 

SuG. — First  divide  both  members  by  a,  and  then  write  the  first  mem- 
ber in  two  parts,  reducing  each  fraction  to  its  lowest  terms.     This  gives 

1 =  c  A-—.     Drop  V  from  each  member  and  -  =  c.      . • .  x  :=  -. 

jc        6  6  6  X  c 

The  pupil  should  also  perform  this  by  the  ordinary  process. 


WITH  ONE  UNKNOWN  QUANTITY.         229 

56.   Solve   ^- ~  -L  -—  =  X  -^1. 

SuG. — This  example  is  readily  solved  by  first  clearing  of  fractions,  if 
the  pupil  notices  that  the  terms  involving  x^  and  x^  then  destroy  each 
other.     But  a  much  more  elegant  solution  is  obtained  by  noticing  that 

(2x  +  3)a;       2.T24.3X  ,   ,  1       .     ,  .,  ,•       -u 

■ — —  = ! z=  X  -4-  1 r  '  whence  the  equation  be- 

2a: -fl  2.1- +1  ^  2x  +  1  ^ 

comes  X  -j-  1  —  ^ ■  -\-  —~  =  x  -\-  1.     Dropping  x  -\-l  from  both 

Ax  — j—  1         oX 

members  and  transposing  and  changing  signs ; — -  =  —-,  or  3x  = 

Ax   -j-    1         oX 

20!  +  1.  .  • .  x=l. 

S„a.    ^4^  =  1  +  1.   .-.   4^  =  i.   and.  =  2. 

2x  2    '   X  X  -f  2       X 

58.    Solve   -j-^ ___-=_,    and  verify. 

2x 
SuG.— First  destroy  the  term  — — 

5 

rncii  ^  —  1  X  2  X   5  X  6 

69.    Solve = -' 

X  —  2       07  —  3       X  —  b       X  —  7 

SuG. — Reducing  each    term   to   a  mixed  number  we  have   1   -\- 
^  1 L-  =  1-1 ?—  —  1 1— .     Whence       ^ 


X— 2  X— 3  '    X  — 6  X  — 7  X  — 2 

= •    Reducing  the  terms  in  each  member  sepa- 

X— 3       X— 6       X— 7  ^  ^ 

rately  to  common  denominators  and  adding,  we  get ;r- = 

(x  —  ^ )  (X  —  oj 

Whence  (x  —  2)(x  —  3)  =  (x  —  6)(x  —  7),  or  x-^ 


(x  — 6nx  — 7)" 

—  13x  -f  42  =  x2  —  5x  4-  6,  and  x  =  ^\. 

24:,  ScH.  4. — It  often  happens  that  an  equation  which  involves  the 
second  or  even  higher  powers  of  the  unknown  quantity  is  still,  virtu- 
ally, a  simple  equation,  since  these  terms  destroy  each  other  in  the  re- 
daction. Several  of  the  preceding  examples  afiford  illustrations  of  this 
fact  if  solved  in  the  regular  way. 


230  SIMPLE    EQUATIONS 

60.  What  value  for  x  satisfies  3  —  x  —  2  {x  —  l)(^  +  2) 
=  (^  _  3)(5  _  ^x)  ?  Ans.,  X  =  If 

61.  Solve  the  equation  (^  —  5)  (^  —  2)  —  {x  —  5)  (2^7  — 5) 
+  '(^  +  7)(^  —  2)  =  0?     Ans.,  X  =  2^^ 

62.  Solve   iL±^  -  2(3.  -  4)  +  (3^  -  2)(2.  -  3)^ 

do  o 

x^  —   — ,    and  verify. 
15 

63.  Solve  {x  +  ^){x  —  f )  _  (^  _^  5)(^  __  3)  ^_  |  =  0, 
and  verify  the  result. 

64    Solve  {a  -^  x){h  +  x)  =  (c  +  x){d  +  x). 

cd  —  ah 


Result  X 


b  —  c 


65.    Solve    ^il^=-^-  + 


X —  c        X  — a      X —  h 

ab{a  +  6  — 2c) 


Residt,  X 


a'^  +  ¥  —  ac  —  be 


66.  Solve  (a  +  x){b  ^  x)    —  a(6  +  c)  =  —  +  xK 

Result,  X  =  -r- 

0 

SuG. — Perform  the  multiplication  indicated  in  the  first  member,  and 
write  the  terms  without  clearing  of  fractions;  this  gives  (a  +  b)x  —  ac 

=  -— •     Whence   (a  -f  h)x  =  r =  ^ — -r-^ —     Dividing  by 

0  0  0 

ac 
(a  +  b)  we  have  x  =  —  • 


SIMPLE  EQUATIONS  CONTAINING  RADICALS. 

2S,  Many  equations  containing  radicals  which  involve 
the  unknown  quantity,  though  not  primarily  appearing  as 
simple  equations,  become  so  after  being  freed  of  such  radi- 
cals. 


INVOLVING   RADICALS.  231 

20,  I*roh.  2,   To  free  an  equation  of  Radicals. 
RULE. — The   common    method   is   to    so   transpose   the 

TERMS  THAT  THE  RADICAL,  IF  THERE  IS  BUT  ONE,  OR  THE  MORE 
COMPLEX  RADICAL,  IF  THERE  ARE  SE\'ERAL,  SHALL  CONSTITUTE 
ONE  ME:MBER,  and  THEN  INYOLVE  K\CH  MEMBER  OF  THE  EQUA- 
TION TO  A  POWER  OF  THE  SAME  DEGREE  AS  THE  RADICAL,  If 
A  RADICAL  STILL  REMAINS  REPEAT  THE  PROCESS,  BEING  CAREFUL 
TO  KEEP  THE  MEilBERS  IN  THE  MOST  CONDENSED  FORM  AND 
LOWEST   TERMS. 

Dem. — That  this  process  frees  the  equation  of  the  radical  which 
constitutes  one  of  its  members  is  evident  from  the  fact  that  a  radical 
quantity  is  involved  to  a  power  of  the  same  degree  as  its  indicated  root 
b}^  drojjping  the  root  sign. 

That  the  process  does  not  destroy  the  equaUty  of  the  members  is 
evident  from  the  fact  that  the  like  powers  of  equal  quantities  are  equaL 
Both  members  are  increased  or  decreased  alike. 

EXAMPLES. 

1.  Find  the  value  of  x  in  V^^x  +  16  =12. 

MODEL  SOIiUTipN. 
OPERATION.  -v/^C  -)-  16  =  12. 

4a;  4-  16  =  144. 
4x  =  128. 
X  =  32. 

Explanation. — I  first  square  each  member  of  the  equation.  The 
first  member,  ^-ix  -j-  16,  is  squared  by  dropping  its  radical  sign,  since 
the  square  of  a  square  root  is  the  number  itself.  The  square  of  the 
second  member,  12,  is  144.  This  process  is  equivalent  to  multiplying 
the  fii-st  member  by  \/'Lv-\-  16  and  the  second  by  12,  hence  as 
^^4  X  +  16  is  equal  to  12,  both  members  have  been  increased  ahke. 
[The  equation  being  freed  from  radicals  the  explanation  becomes  the 
same  as  before.  ] 


2.  Solve  \^5x  +  6=4,  and  v6rify. 


232  SIMPLE    EQUATIONS 

3.  Solve  \/lO^+~a  =  7,  and  verify. 


z>      ;.  23 

5 


Veblfication.      VlU  -  Y  -f  3  =  7,  or  ^/iB  +   3  =  7,  or    %/49 

=  7. 


4.  Solve  v^2a;  +  3  +  4=7,  and  verify. 


I     SuG. — First  transpose  tlie  4  and  unite  it  with  the  7  ;    otherwise, 
squaring  will  not  free  the  equation  of  radicals. 


5.  Solve  8  +  V'Sx  +  6  =  14,  and  verify. 

6.  Solve  and  verify  d\/2x  +  6  +  3  =  15. 


Result,    X  =  5. 

7. 

Solve  Vax  -^  'lab  —  a  =  b. 

a^  A-  b^ 
Besult,  X  =  — — — 
a 

1                                           1 

8. 

Solve  Vx  -{-  x-^  —  x  —  ^=  {).           Besult,  x  =  ^■ 

9. 

Solve  ax  -\-  aV'Aax  -\-  x;^  =  ab. 

SuG. — Before  squaring  put  the  equation  in  the  form  N/2aa;  +  x^=< 

b  —  X.     Besult,  X  = — -;-• 

2(a  +  b) 

10.  Given  v  12  +  ?/  —  vy  =  2  to  find  the  value  of  y. 

Queries. — If  both  members  are  squared  as  the  equation  stands,  will 
it  be  freed  from  radicals  ?  Is  the  first  member  a  binomial  or  a  trino- 
mial ?  What  is  its  square  ?  Which  will  give  the  most  simple  result, 
to  square  it  as  it  stands  or  to  transpose  one  of  the  radicals  ?  Which 
one  is  it  best  to  transpose  ?  Will  once  squaring  free  it  from  radicals  ? 
12  +  2/  =  4  +  4:Vy  +  y,  or  'i*/y  =  8  .•.?/=  4. 

11.  Given  \/x —  16  =  8  —  Vx  to  solve  and  Yerifj. 

SuG. — Solve  this  and  the  five  following  like  the  preceding  by  squar- 
ing twice. 


INVOLVING  RADICALS.  233 

12.  Solve  \/a  -\-  x  -r  Va  —  x  =■   Vax. 

Result,  X  = 


a-  +  4 

13.  Solve  \^x  —  a  =   \/ x  —  g^^- 

lo 

14.  Solve  v^5  X  v^o;  +  2  =   ^Vx  +  2. 

9 
Result^  ^  :=  — . 
JO 

15.  Solve  a  +  a;  =   ^a^  +  xVb^  +    ^ 

Result,    57=    ;; 

4a 

16.  Solve  "iVb   +   j;  —  v/4a  +  a;  ==   \/^. 

(5  — a)2 

Result,  X  =  -T 7— 

, 2  a  — b 

Query. — Why  is  it  best  to  transpose  one  of  tlie  terms  of  the  first 
member  to  the  second  before  squaring  ? 

, a 

17.Given^+ Va  —  x  =      .  to  find  x. 

V  a  —  X 
ScH.  1.— It  is  frequently  best  to  clear  the  equation  of  fractions  first, 
even  when  it  involves  radicals,  especially  if  it  is  noticed  that  the  de- 
nominator is  of  such  a  form  as  will  not  make  the  equation  comph- 
cated. 

In  this  case,  clearing  of  fractions  we  have  jcv/a  —  x-j-^  —  x  =  a. 
Whence  x^a  —  x  =  x,  or  ^/a  —  ic  =  1.     Finally  x  =  a  —  1. 

X  —  ax        v  X 


18.  Solve 


Vx  ^ 


SuG. — This  may  be  solved  in  several  ways.  The  pupil  should  exer- 
cise his  ingenuity  in  seeing  in  what  different  ways  he  can  solve  such 
examples,  and  notice  the  most  elegant  methods.  For  example,  com- 
pare the  following  methods. 

1st  Method.  — Clearing  of  fractions   x-  —  ax'^  =  x.      Dividing  by  x, 

1 

x  —  ax  =  l,  .' .   x  =  - . 


234  SIMPLE  EQUATIONS 

2nd  Method.  —Multiplying  both  members  by  ^x  we  have  x  —  ax  = 

-  =  1,  etc. 
a; 

3rd  Method. — Dividing  numerator  and  denominator  of  the  second 

member  by  ^x,  we  have  ^r-  =  — =•      Whence    multiplying  both 

members  by  ^/x,  x  —  ax  =  1,  etc. 

4.th  Method.— Divide  the  numerator  of  the  first  member  by  its  de- 
nominator and  x/ic  —  a^x  =  — -.     Dividing  both  members  by  x/a;, 

X 

1  —  a  =  -,  or  X  =  :; . 

X  1  —  a 

^^    ^  ,       \/ax  —  b        Svax  —  26 

19.  Solve  -—=: =  — —=z . 

vax  +  b        dvax  +  56 

SuG.  — Clearing  of  fractions  Sax  -\-  26'v/a«  —  5&2  =  Sax  -j-  b\/ax  —  2&2. 
Transposing,  uniting  terms,  and  dividing  by  b,  "J ax  =  3b.     .  • ,  x  = 

96- 

— .     A  much  more  elegant  solution  is  to  reduce  each  member  to  a  mixed 
a 

number,  obtaining  1 — = —  =1 — = .     Dropping  the  1, 

V  ax  -j-  &  3  V  ax  -|-  5b 

2  7 

anddividingby— 6,  — r=: =  — 7== •      Clearing    of    fractions 

V  ax  -j-  >         3  V  ax  -j-  56 
Gx/^-f  106  =  7v/ax"+ 76.     .  • .  n/^=  36,  etc. 
In  a  similar  manner  solve  the  two  following  : 

20.  Solve  —  =  ——^ — .  Itemlt,  ar  =  4. 

V  ^  +  4  V  ^  -f  6 

0-1    a  1      v/^  +  2a        y^  +  4a         ^,      ,,  /_«LV 

21.  Solve  — 73: =  -71= .       llemlt,  x  =  ( r ) 

^x^b        v/a;  +  36  Va— 6/- 

00    a  1         3^  —  1  .    ,    s/^x  —  1 

'22.  Solve  -— == =  1  H ^r 

^/^x  +  1  2 

Sua. — Observing  that  — = =  v^3x  —  1,  we  have  -vSx  —  1  =s 

_  -JSx  +  1 

l_f_2:^_^,  or  V3^=U,  or>v/3^=3.     .•.x  =  3. 
23.  Solve  4^  =  5v/5  -  8  +  ^. 


INTOLVING  RADICALS.  235 

SuG. — In  this  and  the  two  following  use  the  same  expedient  as  in 

the  22nd.    Thus  V^x  —  2  =  5^x  —  8-\-  i^x,  or  54^x  =  6,  or  ^/^  = 
12  _144 

24.  Solve  -^^^  =.  ^^^:^  ^  2v/a. 

SxjG.— We  have   y/x  —  \/a  =  iv^4-  Ifv^oi  or  |n/x  =  2|ya. 
.• .  X  =  16a. 

^^    c^  1         «'^  —  ^'^  ,    ^«^  —  ^ 

25.  Solve  — =: =  c  H 

Vax  +  6  c 


Result,  x=  -(h  A ) . 

a\        c — 1/ 


va  -\-x  -\-  va  —  X  , 

26.  Solve  "7= 7==^  =  v  ^• 

V  a-j-o; —  V  a — x 


Sua. — Eationalize  the  denominator  {224-),  and  after  reducing  some- 
what \/a-3  —  x'^  =  x\/7ii  —  a.  Squaring  a-  —  x^^  mx'^  —  ^aVtnx  -j-  a^ 

or  —  X  ^=  mx  —  2a'v/m.  .  • .  x  = 


2a's/'in 


l-\-m 

—               \/,^  +  a  +  y^r  _,      ,^            a((?  —  l)^ 

27.  Solve    ,         7=  =  c.  Result,  x=-^ — '-. 

^^    „  ,      v/i^Tfl  +  "iVx        ^  „      ,,           4 

28.  Solve Trr  =  9.  Result,  x=-. 

V4.x-\-i  —  'ivx  y 


29.  Solve  v^^  +  v/^  —  v^^  _  v/^  =  tvr~^- 

Sua. — ^Multiply  both  members  by  ^x4-  v'^x,  and  after  some  reduo* 
tion  \/x2 — x=x — Wx.     Squaring,  transposing,  and  combining, 
/-        5  25 


30.  Solve    Vv'^r  +  S—  Vv^^r— 3  =  ^"^^x. 

Result,  j;  =  9. 


236  SIMPLE   EQUATIONS 

31.  SolveJ^  +  J^=J- 


—  x^ 


46c 
Sue. — Squaring   each    member,  notioe    that   the  term 


6  —  c 

4 

X     a       ^a-^       yia^x^      x* 


J  a2  —  x'i 
occurs  on  both  sides. 


82. 

X     a        ^a^    ■   \a^x^      x* 

Result^  X  =  2a. 

33.  Solve  .15^7  +  1.575  —  .875a;  =  .0625x. 

Besulf,  X  =  2. 

34.  Solve   1.2x  —  ,  =  Ax  +  8.9. 

.5 

Besult,  X  =  20. 

35.  Solve  4.8^  —  —      ,  =  1.6a7  +  8.9. 

.5 

Besulf,  X  =  5, 


SUMMARY  OF  PRACTICAL  SUGGESTIONS. 

27* — In  attempting  to  solve  a  simple  equation  which  does  not  con- 
tain radicals,  consider, 

1.  Whether  it  is  best  to  clear  of  fractions  first. 

2.  Look  out  for  compound  negative  terms. 

3.  If  the  numerators  are  polynomials  and  the  denominators  mono- 
mials, it  is  often  better  to  separate  the  fractions  into  parts. 

III. — Ex.  22,  Art.  22,  may  be  written  thus,  by  this  expedient : 
A  X  +i-^\x  +  ^  =  \x-l-  2f.  Whence  (-,\  -  ^  -  \)x  = 
-3|-i-i  or-iix  =  -f|.  .-.  05  =  35. 

4.  It  is  often  expedient,  when  some  of  the  denominators  are  mono- 
mial or  simple,  and  others  polynomial  or  more  complex,  to  clear  of  the 


APPLICATIONS.  237 

most  simple  first,  and  after  each  step  see  that  by  transposition,  uniting 
terms,  etc.,  the  equation  is  kept  in  as  simple  a  form  as  possible.  (See 
Examples  40  to  44. ) 

5.  It  is  sometimes  best  to  transpose  and  unite  some  of  the  terms  be- 
fore clearing  of  fractions.     (See  examples  46  to  54.) 

6.  Be  constantly  on  the  lookout  for  a  factor  which  can  be  divided 
out  of  both  members  of  the  equation  {Ejc.  55),  or  for  terms  which  de- 
stroy each  other  {Ex.  60  to  66). 

7.  It  sometimes  happens  that  by  reducing  fractions  to  mixed  num- 
bers the  terms  wiU  unite  or  destroy  each  other,  especially  when  there 
are  several  polynomial  denominators.     (See  Ex.  59. ) 

28.    When  the  Equation  contains  Radicals,  considee, 

1.  If  there  is  but  one  radical,  by  causing  it  to  constitute  one  mem- 
ber and  the  rational  terms  the  other,  the  equation  can  be  freed  by  in- 
volving both  members  to  the  power  denoted  by  the  index  of  the  radi- 
cal.    (See  Exs.  1  to  9.) 

2.  If  there  are  two  radicals  and  other  terms,  make  the  more  com- 
plex radical  constitute  one  member,  alone, before  squaring.  Such  cases 
usually  require  two  involutions.     (See  Exs.  10  to  16.) 

3.  K  there  is  a  radical  denominator  and  radicals  of  a  similar  form  in 
the  numerators  or  constituting  other  terms,  it  may  be  best  to  clear  of 
fractions  first,  either  in  whole  or  part.     (See  Exs.  17,  18,  29.) 

4.  Frequently  a  fraction  may  be  reduced  to  a  whole  or  mixed  num- 
ber -vs-ith  advantage.     (See  Exs.  19  to  25.) 

5.  It  is  sometimes  best  to  rationahze  a  radical  denominator.  (See 
Exs.  26  to  28.) 

6.  Always  watch  for  an  opportunity  to  divide  out  a  factor,  or  drop 
terms  which  destroy  each  other. 


APPLICATIONS. 

29.  According  to  the  definition  {2),  Algebra  treats  of, 
1st,  The  nature  and  properties  of  the  Equation  ;  and  2nd, 
the  method  of  using  it  as  an  instrument  for  mathematical 
investigation. 

The  Simple  Equation  has  been  so  thoroughly  discussed 


238  SIMPLE    EQUATIONS  : 

in  the  preceding  part  of  the  section,  that  the  pupil  will  now 
be  able  to  use  it  in  solving  problems. 

50,  The  Algebraic  Solutiofi  of  a  problem  con- 
sists of  two  parts : 

1st.  The  Statement,  which  consists  in  expressing  by 
one  or  more  equations  the  conditions  of  the  problem. 

2nd.  The  Solution  of  these  equations  so  as  to  find 
the  values  of  the  unknown  quantities  in  known  ones.  This 
process  has  been  explained,  in  the  case  of  Simple  Equa- 
tions, in  the  preceding  articles. 

51,  The  Statement  of  a  problem  requires  some 
knowledge  of  the  subject  about  which  the  question  is 
asked.  Often  it  requires  a  great  deal  of  this  kind  of  know- 
ledge in  order  to  "  state  a  problem."  This  is  not  Algebra; 
but  it  is  knowledge  which  it  is  more  or  less  important  to 
have  according  to  the  nature  of  the  subject. 

52,  Directions  to  guide  the  student  in  the  State- 
ment of  Problems  : 

1st.  Study  the  meaning  of  the  problem,  so  that,  if  you  had  the  answer 
given,  you  could  prove  it,  noticing  carefully  just  what  operations  you 
would  have  to  perform  upon  the  answer  in  proving.  This  is  called, 
Discovering  the  relatioris  between  the  quantities  involved. 

2nd.  Represent  the  unknown  (required)  quantities  (the  answer)  by 
some  one  or  more  of  the  final  letters  of  the  alphabet,  as  x,  y,  z,  or  to, 
and  the  known  quantities  by  the  other  letters,  or,  as  given  in  the 
problem. 

3rd.  Lastly,  by  combining  the  quantities  involved,  hoth  known  and 
unknown,  according  to  the  conditions  given  in  the  problem  (as  you 
would  to  prove  it,  if  the  answer  were  known)  express  these  relations 
in  the  form  of  an  equation. 

PROBLEMS. 

1 .  What  number  is  that  to  the  double  of  which,  if  18  be 
added,  the  sum  is  82  ? 


APPLICATIONS.  239 

MODEL   SOLUTION. 

Statement. — Let  ic  represent  the  unknown  number  sought.  Then 
the  problem  says  that  double  this  number,  that  is  2a*,  with  18  added, 
that  is  2x  -j-  18,  is  82.     Hence  2x  +  18  =  82  is  the  statement. 

Resolution  or  the  Equation. — [With  this  the  pupil  is  famihar.] 
Solving  2x  +  18  =  82,  we  find  x  =  32.     32  is,  therefore,  the  number 
Bought. 
,    Veeitication.     2  X  32  -}-  18  =  82. 

[Note. — This  is  a  very  simple  question  ;  but  the  pupil  will  do  well 
to  study  it  carefully.  There  are  three  numbers  involved,  the  18,  82, 
and  the  one  to  be  found,  which  we  call  x.  Having  noticed  this,  we  see 
that  the  relations  between  these  numbers  are  explicitly  stated  in  the 
problem,  and  the  statement  is  easily  made.  This  is  not  always  so. 
These  relations  are  often  the  most  difficult  thing  to  discover,  and  their 
discovery  requires  a  knowledge  of  other  subjects  than  Algebra.  Sup- 
pose the  problem  to  be  :  Given  three  masses  of  metal  of  equal  weight, 
one  of  pure  gold,  one  of  pure  silver,  and  one  part  gold  and  part  silver. 
When  they  are  immersed  in  water,  the  gold  displaces  5  ounces,  the  sil- 
ver 9  ounces,  and  the  compound  6  ounces.  ^\'Tiat  part  of  the  last  was 
gold  and  what  part  silver  ?  Now  in  this  problem  the  relations  between 
the  quantities  are  not  explicitly  stated.  Yet  by  a  knowledge  of  Natu- 
ral Philosophy  and  a  httle  of  Mensuration,  they  can  be  found  out,  and 
the  statement  of  the  problem  made  in  an  equation.  ] 

2.  What  number  is  that  to  the  double  of  which,  if  44  be 

added,  the  sum  is  4  times  the  required  number? 

SuG. — What  are  the  numbers  involved  in  this  problem  ?  Ans.  Only 
two,  44,  and  the  number  sought.  Suppose  I  guess  that  the  number 
sought  is  30,  how  will  you  tell  whether  I  am  right  or  not  ?  You  say : 
"  Double  30  and  add  44  to  it,  and,  if  you  are  right,  the  sum  will  equal 
4  times  30."  But  2  X  30  -f  44  does  not  =  4  X  30  ;  so  I  am  wrong. 
Now,  call  the  number  sought  x,  and  use  it  as  30  was  used  in  proving 
that  it  is  not  the  answer,  and  the  statement,  2x  -f-  44  =  4x  is  obtained. 
Whence  x  =  22.     Verify  it. 

3.  What  number  is  that,  the  double  of  which  exceeds  its 
half  by  6?  Ans.,  4. 

4.  The  property  of  two  persons  is  $16000,  and  one  owns 
three  times  as  much  as  the  other.  How  much  has 
each? 


240  SIMPLE  equations: 

Statement.  Let  x  =  the  less  amount. 

Then  3a;  =  the  greater  amount. 
And3x4-a;  =  16000.    .-.    x  =  4000  and  3x  =  12000. 

5.  A  man  being  asked  his  age  replied:  "If  to  my  age  you 
add  its  half,  and  third,  and  then  deduct  10,  the  result 
is  100."     What  was  his  age?  Ans.,  60. 

6.  After  paying  j  of  a  bill  and  -J  of  it,  $92  still  remained 

due.     What  was  the  bill  at  first  ? 

Statement. — Let  x  =  the  amount  of  the  bill;  then  \x  and  ^x  had 
been  paid.  Taking  these  amounts  from  the  bill  we  have  x  —  \x  —  ^x. 
But  this,  by  the  problem,  was  $92.  Hence  x  —  \x  —  -^x  =  92.  .  • .  x  = 
140. 

7.  The  sum  of  2  numbers  is  180  and  the  difference  10. 
What  are  the  numbers  ? 

Statement. — Let  x  =  the  less  number:  then  x  -f-  10  is  the  greater, 
and  X  -{-x  -{-  10  =  180.  .  • .  x  =  85,  and  x  -{-10  ■■=  95,  and  the  two 
numbers  85  and  95. 

8.  The  sum  of  two  numbers  is  5760,  and  the  difference  is 
^  the  greater.     What  are  they  ? 

Statement. — Let  x  =  the  greater;  then  5760  —  x  is  the  less,  and 

X  —  (5760  —  x)  =  |.     .  • .  X  =  3456,  and  the  less  is  2304. 
o 

9.  A  man  divided  80  cents  among  four  beggars;  to  the 
first  two  he  gave  equal  amounts,  to  the  third  twice  as 
much  as  to  one  of  these,  and  to  the  fourth  twice  as 
much  as  to  the  third.    How  much  did  he  give  to  each  ? 

Ans. ,  The  equation  is  x  -\-  x  -{-  2x  -\-  4:X  =  80. 
Whence  he  gave  the  first  and  second  10c. 
each,   the   third   20c.,  and   the  fourth  40 

cents. 

10.  A,  B,  C,  and  D,  invest  |4755  in  a  speculation.     B  fur- 
nishes 3  times  as  much  as  A,  C  as  much  as  A  and 


APPLICATIONS.  241 

B  together,  and  D  as  much  as  C  and  B.     How  much 
does  each  invest  ? 

Ans.,  A,  $317;  B,  $951;  C,$1268;  and  D,  $2219. 


THE  SAME  PROBLEMS  WITH  THE  LITERAL  NOTATION. 

11.  What  niynbef  is  that  to  n  times  which,  if  m  be  added, 
the  sum  is  s  ? 

—  m 


The  equation  is  nx  -}-  m  =  s.  Ans., 


n 


ScH. — To  make  this  conform  to  Pbob.  1,  we  call  s  =  82,m  =  18, 

-  -  s  —  m       82—18       oo-Di-i-v,  ^  —  ■^ 

and  n  =z2.    .  • .     = =  32.     But  the  answer  ap- 

ji  2  n 

plies  just  as  well  to  any  other  problem  of  the  kind,  no  matter  what 

numbers  are  involved.     Thus,  let  the  problem  be, — What  number  is 

that  to  5  times  which,  if  20  be  added,  the  sum  is  100  ?    Now  s  =  100, 

m  =  20,  and  n  =  5.     Whence  = = =  16.     We,  there- 

fore,  see  that  is  a  general  answer  to  all  such  problems. 

12.  What  number  is  that  to  a  times  which,  if  6  be  added, 

the  sum  is  c  times  the  number  ? 

Ans.f  The  equation  is  ax  -^  b  ==  ex,  and  the  num- 
ber  

c  —  a 

Queries.— How  is  this  adapted  to  Prob.  2  ?     What  other  problems 
can  you  state  to  which  this  value  of  x  affords  an  answer  ? 

13.  What  number  is  that,  a  times  which  exceeds  -  times  it 

0 

by  m? 

MiQuation,     ax  —  -  =  m.    .* .  Ans.,  x  =  -; 7 

0  ab  —  1 

is  the  number. 


QtrEBiES. — How  is  this  adapted  to  Prob.  3?    What  other  problems 

can*S'ou  state  to  which  — affords  an  answer  ?    Repeat  tiie  same 

ab  —   1 

inquiries  after  each  of  the  seven  following  examples. 


242  SIMPLE  equations: 

14.    The  property  of  two  persons  is  $m,  and  one  owns  n 
times  as  much  as  the  other.     How  much  has  each  ? 

Ans., and 


1  +   ^  1  +   n 

15.  What  number  is  that  to  which  if th  and  -th  of  it- 

m  n 

self  be  added,  and  then  a  deducted,  the  result  is  5  ? 

(a  +  b)mn 


A71S., 


ran  +  77i  +  n 


16.  After  paying  — th  and  -th  of  a  bilL  a  remained  due. 
*-  m  n 

What  was  the  bill  ?  ,  amn 

Ans.,   ' 


mn  —  n  —  m 
17.  The  sum  of  two  numbers  is  s,  and  their  difference  d. 

What  are  the  numbers  ?     ,  s  -\-  d       ,   s  —  d 

Ans. 

33,  Cor. — Observe  that  the  solution  of  this  problem  proves 
the  very  useful  theorem  : 

Having  the  sum  and  the  difference  of  two  quantities,  the 
greater  is  found  by  taking  half  the  sum  of  the  sum  and  differ- 
ence, and  the  less  by  taking  half  the  difference  of  the  sum  and 
difference.      Or,  since   the    above  results   may   be  written 

-  +  77  and  -  —  -5   the  theorem  may  be  stated: 

2       2  2       2  ^ 

I 

Half  the  sum  plus  half  the  difference  of  two  quantities  is  the 
greater  of  the  quantities,  and  half  the  sum  minus  half  the  dif- 
ference is  the  less.    Thus,  the  sum  of  two  numbers  is  20  and 

s       d 
their  difference  12.      What   are   the   numbers  ?    ^  +  -  = 

10  +  6  =  16,  the  greater;  and  |  —  ^  =  10  —  6  =  4,  the 

less. 


APPLICATIONS.  243 

18.  The  sum  of  two  numbers  is  s  and  the  difference  is  -th 

of  the  greater.     "What  are  the  numbers  ? 

„        .  .  .  X         .  ms      s(m — 1) 

Equation,  x  —  (s  —  x)  =  — .     An^., -,  — — -. 

^  ^  '  m  2m  —  1    2m  —  1 

19.  A  man  divided  $a  among  4  beggars  ;  to  the  first  two  he 

gave  equal  amounts,  to  the  third  m  times  as  much  as 

to  each  of  these,  and  to  the  fourth  m  times  as  much  as 

to  the  third.     How  much  did  he  give  to  each  ? 

a 
Ans.,  To  each  of  the  first  two  $-r-^ ; -,  to  the  third 

am  aiw^ 

%-T—. ■. -,  and  to  the  fourth  %-r-, ■. ;• 

^2  -f  m  +  m-'  ^2  +  m  +  m* 

20.  A,  B,  C  and  D  invest  %s  in  a  speculation.  B  furnishes 
m  times  as  much  as  A  ;  C  as  much  as  A  and  B  together, 
and  D  as  much  as  C  and  B  together.  How  much  does 
each  invest  ? 

S  771.9  8(1  +  ^^) 

Alls. J  A  furnishes  o  ,   a —  ;  B,  ^   ,   a — ;  C,     »  ,  ~< » 

'  3  +  4m  '     '  d  +  4m  '     *     3  +  4?7i 

s(l-i-2m) 

and  D,   .>  .  -; 

3  +  4?7i 


21.  At  a  certain  election  943  men  voted,  and  the  candidate 
chosen  had  a  majority  of  65  votes.  How  many  voted 
for  each?    *  ^7is.,  439  and  504. 

22.  A  farmer  has  two  flocks  of  sheep,  each  containing  the 
same  number.  From  one  he  sells  39,  and  from  the 
other  93,  and  then  finds  just  twice  as  many  in  one  flock 
as  in  the  other.     How  many  did  each  flock  originally 

.  contain?  Ans.,  147. 

23.  A  person  spends  -^  of  his  income  in  board  and  lodging, 
■|-  in  clothing,  and  yV  ^^  charity,  and  has  $318  left. 
What  is  his  income  ?  Ans.,  $720. 


214  SIMPLE    EQUATIONS  l 

24.  From  a  bin  of  wheat  -^  was  taken,  and  20  bushels  were 
added.  After  this  i  of  what  was  then  in  the  bin  was 
sold,  and  i  as  much  as  then  remained  +  30  bushels  was 
put  in,  when  it  was  found  that  the  bin  contained  just  j^ 
as  much  as  at  first.     How  much  did  it  contain  at  first? 

SuG.  — After  taking  out  2,  there  remains  ^x,  calling  x  the  amount 
at  first  in  the  bin.  To  this  add  20  bushels  and  there  is  in  the  bin 
ix  -f-  20.  After  selling  i  of  this  f  remains,  or  f  (gx  -j-  20).  i  of  this 
is  ^{ix  +  20).  .  • .  f  (ix  +  20)  4-  |(ix  +  20)  4-  30  =  hx.  Whence 
X  =  560,  the  answer. 

25.  A  person  has  a  hours  at  his  disposal ;  how  far  may  he 
ride  in  a  coach  which  travels  b  miles  per  hour,  and  yet 
have   time  to  return  on   foot  walking  at  c  miles  per 

hour  ?  .  abc 

Ans. 


6  +  c 

26.  After  paying  -th  of  my  money,  and  then  -th  of 
what  remained,  I  had  $a  left.    How  much  had  I  at  first? 

Equation,  x — \x  —  {^x — \x)^  =  a. 

.  amn 

Ans.,  — — -. 

ran  —  m  —  n  +  1 

27.  A  boy,  being  asked  his  age,  replied,  11  years  are  7  years 
more  than  -|  of  my  age.     How  old  was  he  ? 

Statement. — Let  x  =  his  age.     Then  |x  -j-  7  =  11.     .  • .  x  =  10. 

[Note. — This  example  is  taken  from  Stoddard's  Intellectual  Arith- 
metic, page  118,  Ex.  27.  The  whole  series  of  exajiples  in  that  work 
styled  "Algebraic  Questions,"  on  pages  116-140,  will  afford  good  ex- 
ercise for  the  pupil  at  this  stage  of  his  progress  in  Algebra.  It  wiU  be 
well  for  him  to  solve  them,  both  by  the  processes  given  in  the  Arith- 
metic and  by  the  use  of  the  equation.  He  will  thus  see  how  the  equa- 
tion is  an  instrument  of  great  simplicity  and  power  for  the  solution  of 
mathematical  problems.  ] 

28.  A  boy,  being  asked  how  many  sheep  his  father  had, 
replied,  that  40  were  5  less  than  |  of  his  father's  flock. 
How  many  sheep  had  his  father  ?  Ans.,  60. 


APPLICATIONS.  245 

29.  If  A  can  perform  a  piece  of  work  in  10  days,  and  B  in 
8  days,  in  what  time  will  they  perform  it  together  ? 

Statement. — Let  x  =  the  number  of  days.    Then  since  A  can  do 
the  work  m  10  days,  in  1  day  he  will  do  ^^J,  and  in  x  days  —  of  it.     In 

like  manner  B  will  do  -.     Hence  77:  +  o  =  (^11  ^^^  work)  or  1.     This 

fl.  sometimes  troubles  pupils.  But  let  them  consider  that  if  two  of  us 
do  a  piece  of  work,  one  doing  f  and  the  other  j^,  if  the  work  is  all  done, 
the  sum  of  the  parts  done  is  1. 

30.  There  is  a  certain  piece  of  work  which  A  and  B  can 
do  in  8  days ;  but  A  and  C  can  do  it  in  6  days,  or  B 
and  C  in  10  days.  How  long  would  it  take  any  one  of 
them  to  do  it  alone  ?     How  long  if  all  work  ? 

1      1 

SuG.     -  —  5  is  the  difference  between  what  B  and  C  can  do  in  a  day  ; 

and  —  is  the  sum  of  what  they  can  do  in  a  day.     (See  33,) 

Ans.,  A  in  lOif  days,  C  in  14^2-^,  B  in  34f ,  and  aU  in  5^V 

31.  A,  B,  and  C  can  do  a  piece  of  work  in  4  days,  which  A 
alone  can  do  in  12,  and  B  alone  in  8  days.  C  begins 
the  work  alone,  and  is  joined  after  2  days  by  A,  and 
they  work  together  2  days  more.  A  and  C  being  then 
called  o£f,  how  long  will  it  take  B  to  finish  the  work  ? 

Ans.,  5i  days. 

32.  A  performs  |  of  a  piece  of  work  in  4  days  ;  he  then 
receives  the  assistance  of  B,  and  the  two  together  finish 
it  in  6  days.  Required  the  time  in  which  each  could 
have  done  it  alone  ? 

SuG. — How  much  does  A  do  in  1  day?    Let  x  =  the  time  B  would 

6,6       5 
require  to  do  it  all.     The  equation  is  —  4-  -  =  7;.     A  could  do  it  in 

14       X       7 
14,  and  B  in  21  days. 

33.  A  vessel  can  be  emptied  by  3  taj)s  ;  by  the  first  alone 
it  could  be  emptied  in  80  minutes,  by  the  second  in 


246  SIMPLE    EQUATIONS  : 

200  minutes,  and  by  the  third  in  5  hours.  In  what 
time  will  it  be  emptied  if  all  the  taps  are  opened  at 
once?  ^ns.,  48  minutes. 

34,  A  can  do  a  piece  of  work  in  a  days ;  B  in  6  days,  and  C 

the  same  piece  in  c  days.     In  how  many  days  will  they 

finish  it,  when  all  work  together  ? 

,  abc 

Ans.,  — ; . 

aO'i-oc-\-  ac 

35.  Three  men  A,  B,  and  C  are  employed  on  a  certain  piece 

of  work.  A  and  B  can  do  it  in  «  days ;  A  and  C  in  ^ 
days,  and  B  and  C  in  c  days.  How  long  would  it  take 
each  to  do  it  alone  ?  How  long  would  it  take  them  to 
do  the  work  together  ? 

.          ,   .               2ahc  ^         ^ .  2abc 

Ans.,  A  m 7  days,  13  m 


bo  -\-  ac  —  ab  be  -\-  ab  —  ac 

C   in   — '■ — r->     and    aU    together    in 

ab  -\-  ac  —  be 

2abc 


ab  -}-  ac  -{-  be 


Sua. — Let  the  student  notice  the  singular  symmetry  of  these  an- 
swers. Such  symmetry  is  common,  and  sometimes  becomes  very  use- 
ful in  complicated  processes. 

See  Stoddard's  Intellectual  Arithmetic,  Lesson  LI.,  for  other  exam- 
ples of  this  kind. 


34-,  ScH. — It  is  not  always  expedient  to  use  x  to  represent  the  num- 
ber sought.  The  solution  is  often  simplified  by  letting  x  be  taken  for 
some  number  from  which  the  one  sought  is  readily  found,  or  by  let- 
ting 2x,  3x,  or  some  multiple  of  x  stand  for  the  unknown  quantity. 
The  latter  expedient  is  often  used  to  avoid  fractions. 

36.  There  is  a  fish  whose  head  is  9  inches  long;  the  tail  is 
as  long  as  the  head  and  half  the  body,  and  the  body  is 
as  long  as  the  head  and  tail  together.  "What  is  the 
length  of  the  fish  ?  Ans.,  72  in. 


APPLICATIONS.  247 

SuG.  —Let  X  =  the  length  of  the  body  ;  then  %  -{-  \x=  the  length 
of  the  tail.     The  equation  is  x=  18  -\-\x.     K  x  =  the  whole  length 

the  equation  is9-f-4-9-|-j=a^- 

37.  A  general  whose  cavalry  was  ^  of  his  infantry,  after  a 
defeat  found  that  before  the  battle  -^  of  his  infantry 
less  120,  and  y^  c>f  his  cavah-y  plus  120  had  deserted. 
After  the  battle  he  found  \  of  his  whole  army  in  gar- 
rison, f  on  the  field,  and  that  the  rest  of  those  engaged 
were  either  taken  prisoners  or  slain.  Now  300  plus 
the  number  slain  was  \  the  infantry  he  had  at  first. 
Of  how  many  did  his  army  consist  originally  ? 

SuG. — Let  X  ==  the  number  of  cavalry;  then  3x  =  the  infantry,  and 
4x  =  the  whole  army. 

Alls.,   Whole  army  3600  men;  viz.,  900  cavalry, 

and  2700  infantry. 

38.  A  shepherd,  in  time  of  war,  was  plundered  by  a  party 
of  soldiers,  who  took  i  of  his  flock  and  ^  of  a  sheep 
more;  another  party  took  ^  of  the  remainder  and  ^  of 
a  sheep ;  and  a  third  party  took  ^  of  the  last  remain- 
der and  ^  of  a  sheep,  when  he  had  but  25  sheep  left. 
How  many  had  he  at  first  ? 

SuG.— Letting  12x  =  the  number  in  his  flock  at  first,  9x  —  ^  is  what 
remained  after  the  first  plundering,  6x  —  ^  after  the  second,  and  3x  —  f 
after  the  third.  The  equation  is  3x  — f  =  25.  .-.  12x  =  103,  the 
number  in  his  flock  at  first.  The  pupil  should  notice  that  he  wants 
the  value  of  12x  instead  of  x. 

39.  A  cask.  A,  contains  12  gallons  of  wine  mixed  with  18 

gallons  of  water ;  another,  B,  contains  9  gallons  of 
wine  mixed  with  3  gallons  of  water.  How  many  gal- 
lons must  be  drawn  from  each  to  make  a  mixture  of  7 
gallons  of  wine  and  7  of  water  ? 

SuG.  f  of  the  mixture  in  A  is  wine,  and  f  water;  and  in  B  f  is  wine 
and  \  water.  Let  x  =  the  number  of  gallons  to  be  drawn  from  A ; 
then  14  —  x  represents  the  number  to  be  drawn  from  B.  Of  the  first, 
fx  is  wine,  and  |-x  is  water;  of  the  second  |(14  —  x)  is  wine,  and 
^(14  —  x)  is  water.  But  in  this  new  mixture  the  wine  and  water  are 
equal.   •.•.^x  +  a(14  — x)  =  |x4-4(U  — «). 


248  SIMPLE   EQUATIONS  : 

40.  In  the  composition  of  a  quantity  of  gunpowder  the  ni- 

tre was  10  lbs.  more  than  f  of  the  whole,  the  sulphur 
was  4|-  lbs.  less  than  ^  of  the  whole,  and  the  charcoal 
21bs.  less  than  i  of  the  nitre.  "What  was  the  amount 
of  the  gunpowder  ?  Ans.,  69  lbs. 

SuG.     X  being  the  whole,  the  nitre  is  |ic  +  IO5  the  sulphur  ^a;  —  4:|-, 
and  the  charcoal  |(|x  +  10)  —  2. 

41.  Several  detachments  of  artillery  divided  a  certain  num- 

ber of  cannon  balls.  The  first  took  72,  and  9  of  the 
remainder;  the  next  144,  and  9  of  the  remainder;  the 
third  216,  and  i  of  the  remainder;  the  fourth  288,  and 
9  of  those  that  were  left ;  and  so  on  ;  when  it  was 
found  that  the  balls  had  been  equally  divided.  "What 
was  the  number  of  balls  and  detachments  ? 

Ans.,  4608  balls,  and  8  detachments. 

42.  A  gentleman  bequeathed  his  property  as  follows:  To  his 

eldest  child  he  left  $1800,  and  ^  of  the  rest  of  his 
property;  to  the  second,  twice  that  sum  and  ^  of  what 
then  remained;  to  the  third,  three  times  the  same  sum 
and  -^  of  the  remainder,  and  so  on;  and  by  this  ar- 
rangement his  property  was  divided  equally  among  his 
children.  How  many  children  were  there,  and  what 
was  the  fortune  of  each  ? 

Ans.,  5,  and  $9000  the  fortune  of  each. 

43.  A  brandy  merchant  has  two  kinds  of  brandy  ;  the  one 

cost  y  dollars  per  gallon,  the  other  o".  He  wishes  to 
mix  both  sorts  together  in  such  quantities  that  he  may 
have  50  gallons,  and  each  gallon,  without  profit  or 
loss,  may  be  sold  for  8  dollars.  How  much  must  he 
take  of  each  sort  to  make  up  this  mixture  ? 

Ans.,  37^  gallons  of  the  best,  12|^  of  the  other. 


/- 


APPLICATIOi;iS.  249 

44.  Let  the  price  of  the  best  brandy  in  the  preceding  prob- 
lem =  a  dollars,  the  price  of  the  poorest  =  b  dollars, 
the  number  of  gallons  in  the  mixture  =  n,  and  the 
price  of  the  mixture  =  c.  How  many  gallons  of 
each  kind  must  he  use  ? 

Ans., 7—  gallons  of  the  poorest,  and 

-^ : —  of  the  other. 

a  —  b 

45.  A  father,  who  has  three  children,  bequeaths  his  prop- 
erty by  will  in  the  following  manner:  To  the  eldest 
son  he  leaves  $1,000,  together  with  the  4th  part  of  what 
remains;  to  the  second  he  leaves  $2,000,  together  with 
the  4th  part  of  what  remains  after  the  portion  of  the 
eldest  and  $2,000  have  been  subtracted  from  the  estate; 
to  the  third  he  leaves  $3,000,  together  with  the  4th 
part  of  what  remains  after  the  portions  of  the  two 
other  sons  and  $3,000  have  been  subtracted.  The 
property  is  found  to  be  entirely  disposed  of  by  this  ar- 
rangement.    What  was  the  amount  of  the  property  ? 

Ans.,  $9,000. 

46.  A  father,  who  has  three  children,  bequeaths  his  prop- 

erty by  will  in  the  following  manner :  To  the  eldest 
son  he  leaves  a  sum  a,  together  with  the  nth  part  of 
what  remains ;  to  the  second  he  leaves  a  sum  2a,  to- 
gether with  the  nth  part  of  what  remains  after  the 
portion  of  the  eldest  and  2a  have  been  subtracted 
from  the  estate;  to  the  third  he  leaves  a  sum  3a,  to- 
gether with  the  nth  part  of  what  remains  after  the 
portions  of  the  two  other  sons  and  Sa  have  been  sub- 
tracted. The  property  is  found  to  be  entirely  disposed 
of  by  this  arrangement.  What  was  the  amount  of  the 
property?  (6>i2  —  4n  +  l)a. 

^nS., ; ITT — ■ 

(n  —  iy 


250  ,  SIMPLE  EQUATIONS  : 

47.  A  general  arranging  his  troops  in  the  form  of  a  solid 

square,  finds  he  has  21  men  over,  but  attempting  to 
add  1  man  to  each  side  of  the  square,  finds  he  wants 
200  men  to  fill  up  the  square;  required  the  number  of 
men  on  a  side  at  first,  and  the  whole  number  of  troops. 

Remit,  110  and  12121. 

QuEEY. — Can  the  pnpil  show  that  this  problem  is  the  same  as  the 
following  ? 

The  diflference  between  the  squares  of  two  consecutive  numbers  is 
221.     What  are  the  numbers  ? 

48.  Gold  is  19^  times  as  heavy  as  water,  and  silver  10^ 
times.  A  mixed  mass  weighs  4160  ounces,  and  dis- 
places 250  ounces  of  water.  What  proportions  of  gold 
and  silver  does  it  contain? 

Ana.,  Gold,  3377  ounces;  silver,  783  ounces. 


THE  TRANSLATION  OF  EQUATIONS  INTO  PRACTICAL 
PROBLEMS. 

SS.  Cor. — Since  the  statement  of  a  problem  is  expressing 
the  conditions  of  the  problem  in  an  equation,  or  in  equations 
(30),  it  follows,  conversely,  that  an  equation  may  be  considered 
as  the  enunciation  in  algebraic  language  of  the  conditions  of  a 
practical  problem. 

EXAMPLES. 

rrj  /v» 

1.  Form  problems  of  which  -  —  -  =  6  is  the  statement, 

o        5 

SuG. — Any  number  of  problems  may  be  stated  which  will  meet  the 
requirements.  Thus,  ' '  What  number  is  that  whose  third  part  exceeds 
its  fifth  by  6  ?"  Again  ;  "  A  man  being  asked  his  age  said,  '  When  I 
was  3  as  old  as  I  now  am,  I  was  6  years  older  than  when  I  was  -t  as  old 
as  I  am  now;'  what  was  his  age  ?"  Or,  again  ;  "A  man  being  engaged 
to  work  a  certain  number  of  days,  continued  in  the  service  i  of  the 
whole  time,  but  during  this  time  was  sick,  and  lost  a  number  of  days 
equal  to  ^  of  the  whole  time,  when  finally  he  had  to  break  the  engage- 
ment, having  actually  worked  but  G  days.  What  was  the  whole  period 
engaged  for  ?" 


APPLICATIONS.  251 

2.  Form  problems  of  whicli  x  +  lO^r  =  66  is  the  statement. 

SuG. — The  pupil  may  say,  "What  number  is  that  to  which  if  10 
times  itself  be  added  the  sum  is  66, "  and  meet  the  requisition.  But  it 
is  urged  that  he  exercise  his  ingenuity  to  frame  more  indirect  enunci- 
ations. Thus,  for  this  example,  the  problem  may  be,  "Charles  has 
his  money  in  cents  and  dimes,  of  each  an  equal  number.  The  whole 
of  his  money  is  66  cents.  How  manj^  copper  and  how  many  silver 
pieces  has  he  ?" 


,.  X        X        ^  X        X         ^        .  X        ^x 

tions:    -+-4-2  =  ^;        4-:^l;    2x  — - 

2       4  3       b  2         5 


3.  Give  similar  translations  of  each  of  the  following  equa- 

-,  X         ^x 

2"-2-T 

=  7;  a:+i^  — 760  +  600=2000;       15(fL±il+5 

X  -\-   X  -{-  4: 

b 

SuG. — One  of  the  above  may  be  enunciated  thus  :  "The  tens  digit 
of  a  certain  number  exceeds  its  units  digit  by  4  ;  and  when  the  num- 
ber is  divided  by  the  sum  of  its  digits  the  quotient  is  7.  What  is  the 
number  ?" 

X        X 

4.  Translate  the  equations  3^  +  ^  =  16000  ;   ^  +  n  +  q  — 

2       o 

10  =  100  ;  x  —  ^—^  =  92;  x-{-x  +  2x  +  4:X=  80, 
5       7 

[Note. — See  examples  4,  5,  6,  and  9,  {32).'\ 

5.  Enunciate  problems  which  will  give  rise  to  the  following 

6^  +  7       2^-— 2         2^—1      2(4j7  +  3) 

equations:    — — -.  =:   — - — ;    -^ — ^  + 

^  15  7a;  — 6  5  x-\-3     ^ 

3 


[Note. — The  following  examples  give  rise  to  equations  found  under 
Art,  26,  The  pupil  should  look  them  up,  and  make  similar  enun- 
ciations of  other  examples.] 

6.  What  number  is  that  which  gives  the  same  quotient 
when  28  is  added  to  its  square  root,  and  the  sum  di- 
vided by  its  squaie  root  -f  4,  as  when  38  is  added  to 
the  same  root,  and  this  sum  divided  by  the  square  root 
of  the  number  -f  6  ? 


252  SIMPLE    EQUATIONS 

7.  A  man  has  a  certain  number  of  square  rods  of  land  ly- 

ing in  a  square  ;  if  12  rods  be  added,  the  whole  being 
kept  in  the  form  of  a  square,  his  plat  is  increased  by 
2  rods  on  a  side.     How  much  land  has  he  ? 

8.  What  number  is  that  to  which  if  12  be  added  its 
square  root  is  increased  by  2  ? 

9.  If  4  times  a  certain  number  be  increased  by  1,  the 
square  root  of  this  sum  +  twice  the  square  root  of  the 
number  itself,  divided  by  the  difference  of  the  same 
quantities  is  9.     What  is  the  number  ? 

10.  There  is  a  certain  number  to  which  if  its  own  square 
root  be  added,  and  also  subtracted,  the  difference  be- 
tween the  square  roots  of  the  results  is  1^  times  the 
square  root  of  the  quotient  of  the  number  divided  by 
the  number  -f  its  square  root.     What  is  the  number? 

11.  What  number  is  that  from  which  if  32  be  subtracted, 
the  square  root  of  the  difference  is  equal  to  the  square 
root  of  the  number  —  ^  the  square  root  of  32  ? 

12.  What  number  is  that  from  which  if  a  be  subtracted, 
the  square  root  of  the  difference  is  equal  to  the  square 
root  of  the  number  —  ^  the  square  root  of  a  ? 


Ans. 


25a 
16 


13.  What  number  is  that  whose  square  root  +  2a,  divided 
by  its  square  root  -f-  b,  equals  its  square  root  +  4a, 
divided  by  its   square   root  +  36? 


.^.j^f 


WITH  TWO   UNKNOWN   QUANTITIES.  263 


SECTION   11. 

Simple,  Simultaneous  Equations  with  two   Unknown 
Quantities. 


DEFINITIONS. 

S6,  The  preceding  problems  have  all  been  solved  by  a 
single  equation  containing  only  one  unknown  quantity.  In 
some  of  them  several  quantities  have  been  sought,  it  is  true, 
but  we  have  managed  to  express  these  quantities  by  the  use 
of  a  single  unknown  quantity,  x.  There  are,  however, 
many  problems  in  which  this  is  not  practicable.  In  such 
problems  there  are  two  or  more  quantities  sought,  and  the 
conditions  are  such  as  to  give  rise  to  two  or  more  equa- 
tions. 

III. — To  make  this  latter  statement  clear,  consider  the  following 
problem :  A  says  to  B,  "  If  ^  of  my  age  were  added  to  f  of  yours,  the  sum 
would  be  19^  years."  But,  says  B  to  A,  "  Iff  of  mine  were  subtracted 
from  \  of  yours,  the  remainder  would  be  IS-j-  years. "  Eequired  their 
ages.  Here  are  two  distinct  quantities  sought;  viz.,  A's  age  and  B's 
age.  Suppose  we  represent  A's  age  by  x,  and  B's  by  y.  Now  notice 
that  there  are  also  two  sets  of  conditions.  1st,  the  statement  which 
A  makes  to  B ;  and,  2nd,  the  statement  which  B  makes  to  A.  Accord- 
ing to  the  1st,  we  have  the  equation  5X  +  f !/  =  19^;  and  according  to 
the  2nd,  ^x  —  ly  =  18^. 

37 >  Tndependent  Equations  are  such  as  express 
different  conditions,  and  neither  can  be  reduced  to  the 
other. 

38.  Simultaneous  Equations  are  those  which 
express  different  conditions  of  the  same  problem,  and  con- 
sequently the  letters  representing  the  unknown  quantities 
signify  the  same  things  in  each.  Any  grouj)  of  such  equa- 
tions are  all  satisfied  by  the  same  values  of  the  unknown 
quantities. 


254  SIMPLE    EQUATIONS 

111. — Thus  in  the  example  above  the  two  equations  ^x  -{-  ^y  =  19^ 
and  ^x  —  ^y  =:18\  are  indepemdent  equations,  since  they  express  dif- 
fereni  conditions,  and  neither  can  be  produced  from  the  other.  But, 
since  these  conditions  are  of  the  same  problem,  so  that  x  in  the  first 
equation  means  the  same  as  a;  in  the  second,  and  y  in  the  first,  the 
same  as  y  in  the  second,  they  are  simultaneous  equations.  It  is  evi- 
dent that  the  true  values  of  x  and  y  must  satisfy,  or  verify,  both  equa- 
tions. If,  however,  we  were  to  write  one  equation  from  one  problem, 
and  one  from  another,  while  they  would  be  independent,  they  would  not 
be  simultaneous  ;  x  and  y  would  not  mean  the  same  things  in  the  first 
equation  as  in  the  second.  In  fact,  the  equations  would  be  so  inde- 
pendent, that  they  would  have  nothing  to  do  with  each  other. 

39,  JEjlhninatiofi  is  the  process  of  producing  from 
a  given  set  of  simultaneous  equations  containing  two  or 
more  unknown  quantities,  a  new  set  of  equations  in  which 
one,  at  least,  of  the  unknown  quantities  shall  not  appear. 
The  quantity  thus  disappearing  is  said  to  be  eliminated. 
(The  word  literally  means  2yuUmg  out  of  doors.  We  use 
it  as  meaning  causing  to  disajDpear. ) 

40.  There  are  Three  3Iet1iods  of  Blimination 
in  most  common  use;  viz.,  by  Compaiuson,  by  Substitution, 
and  by  Addition  or  Subtraction.  There  is  also  a  very 
elegant  method,  by  Undetermined  Multipliers,  which  is 
worthy  of  more  attention  than  it  generally  receives,  but 
which  will  be  reserved  for  the  advanced  course. 

ScH. — ^Any  one  of  these  methods  will  solve  all  problems;  but 
some  problems  are  more  readily  worked  by  one  method  than  by  an- 
other, while  it  is  often  convenient  to  use  several  of  the  methods  in  the 
same  problem,  especially  when  there  are  more  than  two  unknown 
quantities. 


ELIMINATION  BY  COMPARISON. 

4:1,  J^rob.  1,  Having  given  two  independent,  simul- 
taneous, simple  equations  between  two  imbnoivn  quantities,  to 
i^educe  fherefrom  by  comjMrison  a  neiu  equation  containi'ng 
only  one  of  the  unknown  quantities.  ^ 


A 


WITH   TWO   UNKNOWN    QUANTITIES.  255 

[Note. — This  Prob.  might  be  stated  simply — To  eliminate  by  com- 
parison ;  but  it  is  very  important  that  the  statement  be  made  in  full, 
as  above.] 

HULE.    1st. — Find    expressions   for   the   value   of   the 

SAME  UNKNOWN  QUANTITY  FROM  EACH  EQUATION,  IN  TERMS  OF  THE 
OTHER  UNKNOWN  QUANTITY  AND  KNOWN  QUANTITIES. 

•    2nd.  Place  these  two  values  equal  to  each  other,  and 

THE  result  WILL  BE  THE  EQUATION  SOUGHT. 

Dem. — The  first  operations  being  performed  according  to  the  rules 
for  simple  equations  with  one  unknown  quantity,  need  no  further  dem- 
onstration. 

2nd.  Having  formed  expressions  for  the  value  of  the  same  unknown 
quantity  in  both  equations,  since  the  equations  are  simultaneous  this 
unkno^-n  quantity  means  the  same  thing  in  the  two  equations,  and 
hence  the  two  expressions  for  its  value  are  equal,     q.  e.  d. 

ScH. — The  resulting  equation  can  be  solved  by  the  rules  already 
given. 

EXAMPLES 

Of  Independent   Simultaneous   Equations. 

1.  Given  4a7  +  ?/  =  34  and  Aij  -\-  x  =  16  ;  to  find  x  and  y 
and  verify  the  values. 

MODEL  SOLUTION. 

34  — t/ 

•••  ^^— r- 

(2)  4?/  -f  X  =  16  .  • .  X  =  16  —  4?/ 

^1^^  =  16-4,    ...  y  =  2 

4x+2=34  .'.  x  =  8 

Explanation. — Transposing  y  and  dividing  by  4  I  have  from  the  1st 

equation,  x  = '-.     Transposing  4?/  in  the  2nd  equation  I  have 

X  =  16  —  -iy.     Now,  since  these  equations  are  assumed  to  be  simulta- 
neous, X  means  the  same  thing  in  both  ;  and  since  things  that  are  equal 

to  the  same  thing  are  equal  to  each  other,  — - — -  =  16  —  4i/.     From 

this  equation  I  find  y  =  2.     Finally,  since  y  is  iound  to  be  2,  putting 
2  for  y  in  the  first  eotr^tion,  I  havo  -Ix  -f-  ^  =  34  ;  whonce  x  =  8. 


OPEEATION.    (1)  4x  -f-  y  =  34 


256  SIMPLE    EQUATIONS 

VEEincATioN. — Substituting  for  x,  8,  and  for  y,  2,  in  the  1st  equation 
I  have  32  -f  2  =  34;  and  in  the  2nd,  8  -f-  8  =  16,  both  of  which  equa- 
tions are  satisfied. 

2.  Given  5^  +  4?/  ^  58,  and  3^7  +  7^=  67,  to  find  x  and 
y,  eliminating  by  comparison.     Verify  the  results. 

Besults,    -j  „' 

3.  Given  11^  +  3i/  =  100  and  4:X  —  ly  =  4,  to  find  x  and 
y,  eliminating  by  comparison.     Verify  the  results. 

x  =  8, 
4. 


Results,   \ 

\y 


.     r.  ,  .  r.        ^  +  '"5        „        3^  —  2?/       , 

4.  Same  as  above,  given  8 -j—  =7 p and 

8  —  ?/        _.j        2.^  +  1 
4^- -^=241 —. 

SuG. — Observe  the  effect  of  the  —  sign  be^bre  the  compound  quan- 

5  -I-  7x 
tities.     The  value  of  y  from  the   1st  is  y  =  —^ — ,    and  from   the 

160  —  30x      ^^  5  4-  Ix         160  —  30x        ,  _        ^ 

2nd,  V  =  7 •     Hence      — \ = ■ ,  andx  =  5,  and 

'  ^  2  8  2 

5.  Given  ax  -^  hy  ^=  m,  and  ex  -\-  dy=  7i,  to  find  x  and  y, 
eliminating  by  comparison. 

Sua— From  the  1st,  y  =  — ;  and  from  the  2nd,  y  = — . 

m  —  ax       n  —  ex        ^          hn  —  dm      ^^         •     .      -,     ^      ■.    i.-j.  l 
and  a;  =  -^ ;-.     Now,  instead  of  substitut- 


'  '        b  d  be  —  ad 

ing  this  value  of  x  for  x  in  one  of  the  given  equations,  it  will  bo  found 

more  expeditious  to  eliminate  x  as  we  did  y.     Thus,  from  the  1st,  x  = 

TO  —  by      ,  ^      ^i    r>  1        ^  —  (^y         '^  —  ^v     ^^  —  (^y      J 

-:  and  from  the  2nd,  x  =  -.     .  • . = -,    and 

a  c  a  c 

(try  CTTl 

y  =    — — .     To  find  this  value  of  y  by  substituting  for  x  m  the 

ad  —  be 

!>    J.  i.-       -i.        1        bn  —  dm         ,  bn  —  dm- 

first  equation  its  value, -. ,  we  have  a- 4-  by  =  m,    or 

^  be  —  ad  bc  —  ad'^ 

abn  —  adm   ,   ,  _  .  .        ,  abn  —  adm 

-\-by  =  m.      By  transposition,  by  =  m 


be  —  ad  be  —  ad 

bem  —  adm  —  abn  -J- adm       bem  —  abn  ^.  .  -,.      ,     ,  cm  —an 

T 7 =  —r 5-.  Dividing  by  b,y  = —. 

be  —  ad         .  be  —  ad  °  ^       6c  —  ad 

Let  the  pupil  consider  whether  this  is  the  same  as  the  former  value  of 
y ;  and  if  it  is,  how  we  might  have  made  the  results  alike  in  form. 


WITH    TWO    UNKNOWN   QUANTITIES.  257 

ELIMINATION  BY   SUBSTITUTION. 

4:2,  J*rob,  2,  Having  given  two  independent,  simulta 
neous,  simple  equations,  between  two  unhnoivn  quantities,  to  de 
duce  therefrom  by  substitution  a  single  equation  with  but  on> 
of  the  unknown  quantities. 

RULE. — 1st.  Find  from  one  of  the  equations  the  value 

OF  THE  UNKNOWN  QUANTITY  TO  BE  ELIMINATED,  IN  TERMS  OF  THE 
OTHER  UNKNOWN  QUANTITY  AND  KNOWN  QUANTITIES. 

2nd.  Substitute  this  value  for  the  same  unknown  quan- 
tity  IN  THE  OTHER  EQUATION. 

Dem. — The  first  process  consists  in  the  solution  of  a  simple  equation, 
and  is  demonstrated  in  the  same  way. 

The  second  process  is  self  evident,  since,  the  equations  being  simul- 
taneous, the  letters  mean  the  same  thing  in  both,  and  it  does  not  de- 
stroy the  equality  of  the  members  to  replace  any  quantity  by  its  equal. 

Q.  E.  D. 

EXAMPLES. 

X   +    y 


1.  Given  the  independent,  simultaneous  equations 


2 


^        ^  =  8,  and  "^^^  +  "^--^  =  11,  to  find  x  and 


3  '  8         '         4 

t/,  eliminating  by  substitution.     Verify  the  results. 

MODEL    SOLUTION. 

/-iN^  +  v     ^ — y     o 

OPEBATION.  (1)      ^ — ^  =  8 

3x  -f  3?/  —  2x  +  2y  =  48 
jc  -f-  5?/  =  48 
a;  =  48  —  5t/ 

(2)  ^4-^-^  =  11 

,o^    48  -  5.V  4-  y    ,   48  -  Sy  -  y 

(3) +  I =  11 

48-4.V       24-3y_ 

3         "^         2 
96  —  8i/  +  72  —  9y  =  66 
—  17y  =  —  102 

.-.     2/  =  6 


258  SIMPLE    EQUATIONS 

XX 

.-.     ic  =  18 

ErpiANATiON. — Taking  equation  (1)  I  clear  it  of  fractions  and  solve 

it  for  X,  finding  that  x  =  48  —  5y.     Now  I  observe  that  if  I  take  the 

2d  equation  and  substitute  this  value  of  x  for  x,  I  shall  have  a  simple 

equation  with  only  the  unknown  quantity  y  in  it.     This  substitution 

does  not  destroy  the  equality,  since,  the  equations  being  simultaneous, 

X  has   the   same  value  in   both.      Making  the   substitution  I  have 

48  —  5v+    y   ,   48-—  5?/  — V       ,.,      -o  ^     •      .r,-  ..      ,     .^, 
^ — ■ — -  -| Y '~  =  II-    lieducmg  this  equation  by  the 

method    for   simple  equations  with  one    unknown    quantity,  I  find 

2/ =6. 

Finally,  resuming  (1)  I  substitute  this  value  of  y  for  y,  which  evi- 

X  -i-6      X 6 

dently  does  not  destroy  the  equation,  and  have  —J- —  =  8.    Solv- 

2  o 

ing  (4),  which  is  now  a  simple  equation  with  one  unknown  quantity,  I 

find  X  =  18.     (Instead  of  taking  (1)  in  its  first  form  it  would  be  better, 

because  so  much  shorter,   to  take  its  reduced  form,  tc  =  48  —  5y. 

.'.  ic  =  18.) 

[Note. — The  student  should  keep  a  sharp  lookout  for  opportunities 
to  effect  such  reductions  of  terms  as  are  made  in  the  equations  follow- 
ing (3)  and  (4).     In  the  latter  the  process  consists  in  observing  that 

-— —  is  -  -|-  3,  and —  is  —  q  -|-  2,  hence  the  first  member  be- 

comes  J5  +  3  —  o  ~f~  2'  ^^^  transposing  the  3  and  2,  we  have  -  —  x  =  3, 

a  O  2  3 

aU  of  which  can  readily  be  effected  mentally,  ] 

Veeification. — Substituting  in  (1)  18  for  x,  and  6  for  y,  I  have 

-1  Q     I     n  1  Q /» 

— ^ —  =8,  or  12  —  4  =  8.     Also,  in  Uke  manner  from  (2), 

1 Q     I     n  -f  Q   n 

I  have      .7"    H — ■  =11,  or  8  -f  3  =  11.      Whence  I  see  that 

3         '        4  ' 

X  =:  18  and  y  =  6  satisfies  both  equations. 

2.  Given  3^  —  2ij  =  1   and   3y  —  4:X  =  1,   to   solve   as 
above.  Result,  x  =  ^  and'i/  =  7. 


WITH   T^^O    UNKNOWT^    QUANTITIES.  259 

SxjG. — From  the  second,  y  =  —\: — ,  hence  the  first  becomes  3x  — 

o 

"^     =1  .  •  .  X = 5.  Taking  the  reduced  form  of  the  second,  y = — —  , 
3  <J 

and  putting  5  for  x,  y  =  1.     Verify  both  equations. 

Besult,  X  =  12  and  2/  =  6. 

4.  Given  |  + 1  =  1  and  |  + 1  =  1.     Verify. 

Result,  X  =^  —  6  and  y  =  12. 

^  ah  ^  c      d 

5.  Given  --{--  =  m  and  -  +  -  =  ti. 

X      y  X      y 

SuG. — If  we  clear  these  equations  of  fractions  they  will  become 
quite  complex.     But  multiplying  the  first  by  c  and  the  second  by  a, 

ac    ,   be  ,  ac     ,   ad  -^         i.u    ^  ^'^ 

we  have =^cm,  and =an.     From  the  former  —  = 

X  ^  y  X  y  ,  ^, 

be  ac     .  be        ad 

cm ,  which  substituted  in  the  latter  for  —  gives  cin = 

y  X  y        y 

ad  —  he  ad  —  &c       ^,  .         ,         „ 

an,  or =an  —  cm,-.  y= .      This  value  of  y  may 

'  y  ^       an  —  cm 

a      {an  —  cm)b  a 

now  be  substituted  in  the  first,  giving  -  -\ —  =  m,  or  -  =  m 

X  QCl OC  tL 

abn  —  hem  adm  —  6cm  —  abn  +  bcm  a  ((7m  —  bn)  1  

ad  —  be  ad  —  be  ad  —  be    '  x 

^  .• .  x=  — .     To  get  the  value  of  x  from  such  an  equa- 

ad  —  be  dm  —  b7i 

tion  as  -  =: ,  simply  divide  1  by  both  members,  i.  e^  take  the 

X        ad — be 
reciprocals  ;  since  the  same  quantity,  1,  divided  by  equal  quantities 
(the  two  members)  gives  equal  quantities. 

6.  Given h  ^  =  1,  and  -  H =  1. 

X      y  ^      y 


Result,  X  =■  m  -\-  n,  and  y  =  m  -{■  n. 
Result,  X  =  3a,  and  y  =  —  26. 


XV       ^        ^  X       y        2^^.. 
7.  Given-  +  |  =  l.and3^  +  -  =  3.     Venfy 


260  SIMPLE    EQUATIONS 

ELIMINATION  BY  ADDITION  OR  SUBTRACTION. 

43.  I^roh,  3.  Haolng  given  two  independent,  simulta' 
neous,  simple  equations  between  two  unknown  quantities,  to 
deduce  therefrom  by  Addition  or  Subtraction  a  single  equation 
with  but  one  unknown  quantity. 

R  ULE. — 1st.  Beduce  the  equations  to  the  ^oems  ax  +  by 
^==  m,  AND  ex  -\-  dy  =  n. 

2nd.  If  the  coefficients  of  the  quantity  to  be  eliminated 

ARE  NOT  ALIKE  IN  BOTH  EQUATIONS,  MAKE  THEM  SO  BY  FINDING 
THEIR  L.  C.  M.  AND  THEN  MULTIPLYING  EACH  EQUATION  BY  THIS 
L.  C.  M.  EXCLUSIVE  OF  THE  FACTOR  WHICH  THE  TERM  TO  BE  ELIM- 
INATED ALREADY  CONTAINS. 

3rd.  If  the  signs  of  the  terms  containing  the  quantity 
TO  be  eliminated  are  alike  in  both  equations,  subtract  one 
equation  from  the  other,  member  by  member.   If  these 

SIGNS   ARE   unlike,    ADD    THE   EQUATIONS. 

Dem. — The  first  operations  are  performed  according  to  the  rules 
already  given  for  clearing  of  fractions,  transposition,  and  uniting  terms, 
and  hence  do  not  vitiate  the  equations.  The  object  of  this  reduction 
is  to  make  the  two  subsequent  steps  practicable. 

The  second  step  does  not  vitiate  the  equations,  since  in  the  case  of 
either  equation,  both  its  members  are  multiiDlied  by  the  same  number. 

The  3rd  step  ehminates  the  unknown  qiiantity,  since,  as  the  terms 
containing  the  quantity  to  be  ehminated  have  the  same  numerical 
value,  if  they  have  the  same  sign,  by  suUraciing  the  equations  one  "vvill 
destroy  the  other,  and  if  they  have  different  signs,  by  adding  the  equa- 
tions they  wiU  destroy  each  other.  The  result  is  a  true  equation,  since, 
If  equals  (the  two  members  of  one  equation)  are  added  to  equals  (the 
two  members  of  the  other  equation),  the  sums  are  equal.  Thus  we 
have  a  new  equation  with  but  one  unknown  quantity,     q,  e.  d. 

examples. 

,      _,.           x  —  2        10  — a;        ?/  — 10        ^     2v  +  4: 
1.    Given —  =  ■ — : and 


5  3  4  3 

— - —  =  — J — ,    to  eliminate  by  addition  or  subtrac- 
tion and  find  the  values  of  x  and  y.    Verify  the  results. 


WITH   TWO    UNKNOWN   QUANTITIES.  261 


MODEL   SOLUTION. — OPEBATION. 

x-2       10— a!       y  —  10     ^        ^y  +  4       2x -\- y   _  x  -\- IZ 

6                  o                    i            ^"'^             3                   8                   i 

12x— 24-200+200;  =  irjy—150            16y  +  82  —  6x  —  3y  =  6x  -\-  78 

32a;  —  15y  =  74                (4)            12x  —  ldy  =  —  46 

(5)  4:lGx  —  ldoy=       962 

(6)  180.C  —  195?/  =  —  690 
236a;                =     1652 

(1) 

C3) 


X  =  7 
(7)  84  — ISy  =—   46 

—  13y  =— 130 
y  =  10 

Explanation. — Clearing  (1)  of  fractions  and  reducing  to  the  required 
form  I  have  32x  —  15y  ■=^  74.  And  in  Uke  manner  (2;  becomes  12x  — 
13f/  =  —  46.  I  will  eliminate  y.  As  the  L.  C.  M.  of  15  and  13  is  their 
product,  I  multiply  (3)  by  13  and  (4)  by  15,  which  makes  the  terms 
containing  y  of  the  same  numerical  value,  and  does  not  vitiate  the 
equations,  since.  Equals  multiplied  by  the  same  number  give  equal 
products.  I  thus  have  (5),  416x  —  195?/  =  962,  and  (6)  ISOx  —  195y 
=  —  690.  Observing  that  by  subtracting  the  equations  member  from 
member,  —  195:?/  ^^  disappear  from  the  result,  I  subtract  (6)  from  (5) 
and  have  236x  =  1652,  from  which  y  is  eliminated.  This  is  a  true 
equation  deduced  from  (5)  and  (6),  since.  Equals  added  to  equals  give 
equal  sums.  Dividing  by  236  I  find  x  =  7.  Substitutmg  this  value  of 
X  in  (4)  (any  other  of  the  equations  would  answer  but  this  is  the  sim- 
plest), I  have  (7)    84  —  13y  =  —  46.     'VNTience  I  find  y  =  10. 

y 2 

Vebification.- — Putting  7  for  x  and  10  for  y,  (1)  becomes  — - —  — 

5 

20  +  4 


10 

3 

7_ 

10 

— 

10 

4 

14 

±_ 

10 

7 

4- 

13 

1=0.     In  Uke  manner  (2)  becomes 


3  :=  5.     Whence  it  appears  that  x  =  7 


8  4 

and  y  =  10  satisfies  both  equations. 

2.    Given  77a:  —  12?/  =  289,  and  55a;  +  27y  =  491,  to 
find  X  and  y. 

MODEL   solution. 

operation.        (1)       77x  —    12y  =  289       (2)     55x  +    27?/  =  491 
693x  —  108?/  =  2601  220x  +  108?/  =  1964 

220x  +  108?/  =  1964 

(3)      913x  =  4565 


262  SIMPLE   EQUATIONS 

(3)     913x  =  4565 
ic  =  5 
(4)     275  +  lly  =^  491 
21y  =  216 

Explanation. — Since  these  equations  are  of  the  required  form,  I 
have  only  to  make  the  numerical  values  of  the  terms  to  be  ehminated 
alike.  I  will  eUminate  y,  since  its  coefficients  are  smaller  than  those 
of  X  and  the  process  will  not  involve  as  large  numbers.  The  L.  C.  M. 
of  12  and  27  is  108,  hence  I  multiply  (1)  by  9,  obtaining  693x  —  108?/ 
=  2601,  and  (2)  by  4,  obtaining  220a;  +  108?/  ==  1964.  This  process 
does  not  vitiate  the  equations,  since.  Equals  multiphecl  by  the  same 
number  give  equal  products.  I  now  observe  that  the  signs  of  108?/  in 
the  two  equations  are  different,  and  consequently  that  by  adding  the 
equations  these  terms  will  destroy  each  other  and  give  an  equation  in 
X  only.  Adding  gives  a  true  equation,  since,  Equals  added  to  equals 
give  equal  sums.  I  therefore  have  913x  =  4565,  or  a;  =5.  Substitut- 
ing this  result  in  (2)  I  have  275  -|-  27?/  =  491,  whence  y  =  8. 

44.  ScH. — It  is  usually  expedient,  in  examples  involving  two  un- 
known quantities,  to  find  the  value  of  the  second  by  substitution ;  but 
this  is  by  no  means  always  so.  The  pupil  should  perform  the  exam- 
ple in  several  ways,  if  he  can  discern  no  choice  of  ways  at  first,  and 
then  compare  the  methods  with  reference  to  practicability. 


3. 

Given 

,    5.r  +  2v        3v  —  12  +  8.^          ,        15  +  2.r  —  4y 
^+        6                         5              -  ^                   3            ' 

^    7.r4-13  — 5//    ,               ^           3^  +  2?/— IG     ^ 
and -^  X  =  2y  —   : ,    to 

find  X  and  y. 

Results,  a;  =  4,  y  =  5. 

4 

^.        ,    ,  25  +   5v       7^—6        ,^        3^  — 10+7.V 

Given  1  H ^ — —  =10 r-r ~ 

bo                                      1^ 

and  ^^ :: — —  =5^ ^— ^,    to  find  the  values  of  x 

and  y. 

Results,  X  =  o  and  y  =  7. 


WITH   TWO    UNKNOWN    QUANTITIES.  263 

5.    Given  — :=  — -- —  and  8x  —  by  =  1,  to  find 

the  Talues  of  x  and  y. 

Besults,  X  =  7  and  y  =  11. 

45,  ScH. — In  practice  it  often  requires  considerable  discrimination 
to  determine  which  of  the  methods  of  ehmination  to  employ.  But,  as 
any  one  method  will  solve  all  cases,  the  pupil  need  not  hesitate  too 
long  in  attempting  to  select  the  best  one.  K  he  sees  any  reason  why 
one  method  Mill  be  better  than  another  in  the  given  case,  he  -mil  of 
course  use  it;  but,  if  no  such  ground  for  choice  is  apparent,  it  will  often 
be  well  to  try  more  than  one  method,  and  see  if  one  is  any  more  expe- 
ditious than  another. 

exa:\iples  for  geneeal  practice. 

1.  Given  3x  —  5y  =  13  and  2^  +  7y  =  81,  to  find  x  and  y. 

Besults,  X  =  16  and  y  =  1. 


2. 

Given  o  +  7  =  9  ^^^  J  "*"  c  "^  '^'  ^^  ^^*^  ^  ^^^  V- 

Results,  x=12  and  y  =  20. 

3. 

4.x  — ^y-l        3^        2y        5        ,  v— 1        ^ 
^^"^^            5"           ~  10        15        6  ""^  ■   3      +  2 

20        ^-15    ^Q  +  ^Q^^of^ndx^uAy. 

Results,  X  =  3  and  y  =  2. 

2y        8.r  — 2  4.  +  y      x  —  y  ^ 

4.  Given  -— —  =1 ^; \-  — - —    and    Ix  = 

18  db  6  b 

12y,  to  find  x  and  y. 

SuG. — The  first  equation  reduces  at  ouce  to  7.^  =:  7  -f-  H?/-  I^i  this 
case  the  pupil  will  see  that  the  three  methods  of  ehminating  x  are 
almost  identical.  Comparing  the  values  of  Ix,  we  have  l'2y  =  7 -\-lly  ; 
or  we  may  call  it  suhstihding  the  value  of  7x  found  in  the  second  equa- 
tion ;  i.  e.  12y,  for  7x  in  the  first.  Subtracting  the  second  from  the 
first  we  should  have  0=7  —  y. 

r      r.-  15  21  .n  -,    20  6  ^     ^        ^       , 

o.  Given 1 =  10  and =  2,  to  find  x  and  y. 

X        y  X        y 


264  SIMPLE    EQUATIONS 

10       3 

SuG. — The  second  becomes  by  dividing  by  2, =  1,    and  by 

X       y 

multiplying  by  7, =  7.     Now  adding  flais  to  the  first  we  ha\e 

X       y 

—  =:  17,  or  ic  =  5  and  hence  w  =  3. 
X  ^ 

6.  Given  --\ —  =  m   and =  ^,  to  find  x  and  y. 

X     ij  ^     y 

ad        hd         ,  .he        hd        ,  ad  -\- he        ,       ,   , 

SuG. =  dm  and =  bn,  .  • .  =  am  +  bn 

X  y  X         y  X 

and  X  =  - — — — .     Instead  of  substituting  this  value  of  x,  it  will  be 

dm  -\-  on 
less  work  to  eliminate  x  from  the  two  given  equations  as  we  did  y. 

ni  1         «'^     .    ^^  -,  «c         ad  ^        ,  .       ,. 

Thus  we  have =  cm  -and =  an,    and    subtracting, 

X      '     y  X         y 

he,  -f-  nd  he  4-  ad 

=  cm  —  an,  .  • .  w  = • 

y  cm  —  an 

7.  Given  \x -\-  \y  =  14  and  \x  -{-  \y=  11,  to  find  x  and 
y,  and  verify. 

Suggestion. — Do  not  clear  of  fractions. 

8.  Given  x — —— —  =  1  H —r—  and  — 

11  OO  O 

V  — 5       11j?+152       3v  +  1     ^      ^    -,  -,  , 

^L- — = -J—l — ^   to   find  X   and  2/,   and 

4:  1^  J 

verify. 

16  4-60.r       16jtv— 107 


9.  Given  8^- 


3?/  - —  1  5  +  2?/     '  !  to  find  the  values 


and2  +  6y  +  9.=  -^^-^^— , 


of  07  and  2/. 


SuG.  — Multiplying  the  1st  equation  by  5  -j-  2?/,  and  reduce  before 
multiplying  by  3^  —  1.  Clear  the  2nd  of  fractions.  Whence  289  y  — 
340x  =^  187,  and  15x  -f  2t/=  36.   .*.  a;  =  2,  y  =  3. 

AO.  Given  3a:  +  Gv  +  1  = ^ : r^-, 

'AX — 4?/  +  o 

and  3a: -, -i —  =  —z -r--,  to  nnd  the   values 

4y  —  1  6y  —  4 

of  X  and  y.  Result^  x  =  9,  and  ?/  =  2. 

r 


WITH   H\^0   UNKNOWN   QUANTITIES.  265 

128^2  _i8,/2+  217 


11.  Given  16^  -}-  6y  —  1 


Sx  —  '3y  ^  '2 


,  10^  +  10//—  35        _       54  ,     .   ^  ,^ 

and  -- — ,    .  •    ,    ._^     =  5  —  - — — -  -,  to  find  the 

'2jo  -\-  '2y  -\-  6  dx  -\-2y  —  1 

values  of  x  and  y.  Result,  x^=Q,  and  y  =  5. 

12.  Given   (^  +  5)(y  +  7)  =  (^  +  1)(!/ —  9)  +  112,    and 
2a:  +  10  =  3?/  +  1,  to  find  the  values  of  x  and  y. 

Result,  07  =  3,  and  y  =  5. 

13.  Given   if""  ^a7c^-h^)       2b^  ^  ^^  ^^^  ^^^  ^^1" 


I  b^y  -\ = 1-  c^x,  i  ues  ol  x  and  y. 


Result,  X  =  — ,  y  ■ 


be  c 


2 


SuG. — From  the  1st  equation  a;  ^  ^  —  -.     Substituting  tliis  in  the 
2nd,  h-y  -\ = 1"  ;     —  ^^^'       Whence,  transposing    and 

uniting,     —^y    =    -— +     — ^—       or      -^y     = 

a  -4-26      &3 —  (.3  a -I- 26 

— ^^ —  X  — -, ,  and  y  =  — ■ —  .     Substituting  this  value  of  y  in  the 

1st  equation   and  reducing,  we  find  a  =  — .     These  equations  can  be 

solved  by  a  variety  of  methods,  but  the  pupil  should  constanth'  exercise 
his  inventive  genius  to  discover  the  most  .expeditious  and  elegant 
solutions. 

14    Given  2a;  +  0.4y  =  1.2,  and  3.4a7  —  0.02?/  -=  0.01,  to 
find  the  values  of  x  and  y. 

Result,  x  =  .02,y=  2.9. 

IB.  Gi>..    ■  '  +  '  -  "■™ 


la 


56ar  + 13.421^  =  763.4 

58.54 


Result,  i^'^^        rro^/     '  I"  nearly. 


APPLICATIONS. 

1.    There   are  two  numbers,  such,  that  three  times  the 
greater  added  to  one-third  the  less  is  36;  and  if  twice 


266  SIMPLE   EQUATIONS 

the  greater  be  subtracted  from  6  times  the  less,  and 
the  remainder  divided  by  8,  the  quotient  will  be  4. 
"What  are  the  numbers  ? 


OPEEATION. 

Then 


MODEL   SOLUTION. 

Let 

X  =  the  greater  number, 

and 

y  =  the  less  number. 

(1) 

Sx-\-y  = 

36 

(2) 

6y  —  2x 

8 

4 

(3) 

6y  +  54.x  = 

648 

(4) 

6y  —  2x  = 

32 

5&X  = 

616 

X  = 

11 

(5) 

6y  —  22  = 

32 

y  = 

9 

Explanation. — As  there  are  two  unknown  quantities  involved  in  this 
example,  L  e. ,  the  two  numbers  sought,  I  let  x  represent  the  greater 
and  y  the  less.  There  are  also  two  sets  of  conditions  stated  in  the 
problem  :  1st,  3  times  the  greater  added  to  ^-  the  less  is  36.  This,  ac- 
cording to  the  notation,  is  3.i;  -\-  ^y  z=  36,  which  is  the  first  equation. 
The  2d  set  of  conditions  is,  that  tmce  the  greater  is  to  be  subtracted  from 
6  times  the  less,  which  is  6y  —  2.r,  and  this  difference  divided  by  8,  i.  e. 

Qy 2x 

- — -— ^ .     This  quotient  is  equal  to  4.     Hence  the  second  equation, 
8 

6v  —  2x         , 

[Note. — The  explanation  of  the  resolution  of  the  equations  can  be 
given  as  heretofore.  But  as  the  attention  is  now  to  be  directed  chiefly 
to  the  statement  of  questions,  if  the  former  subject  has  been  thoroughly- 
mastered,  little  need  be  said  about  it  here.  Nevertheless  the  pupil  should 
not  he  allowed  to  run  over,  his  solution  simply  reading  the  successive  equa- 
tions. If  he  does  anything  more  than  simply  give,  as  above,  the  ex- 
planation of  the  statement  then  saying  ' '  Solving  which  equations  I 
find  K  =  11  and  t/  =  9,"  he  should  at  least  tell  what  he  does,  thus  : 
"Multiplying  (1)  by  18,  and  (2)  by  8,  I  have  6y  -f-  54a;  =  648.  and 
Qy  —  2x  =  32.  Subtracting  (4)  from  (3),  member  by  member,  I  elimi- 
nate y  and  find  56x  =  616.     Dividing  by  56,  x  =  11,  &c."] 


WITH   TWO    UNKNOWN   QUANTITIES.  267 

2.  Find  two  numbers,  such,  that  if  the  first  be  increased 
by  a,  it  will  be  m  times  the  second;  and  if  the  second 
be  increased  by  h,  it  will  be  n  times  the  first  ? 

a  -\-  hm  b  -\-    an 

EesuU, -,    and  -. 

mn  —  1  mn  —  1 

3.  What  two  numbers  are  those,  to  i  of  the  sum  of  which 
if  I  add  13,  the  result  will  be  17  ;  and  if  from  ^  their 

.  difference   I    subtract   1,    the   remainder   will  be  2? 
Verify.  Ans.,  9  and  3. 

[Note. — In  verifying  the  results  in  such  examples  as  these,  no 
attention  should  be  paid  to  the  equations ;  but  the  results  should  be 
tested  directly  by  the  statement.  Thus,  in  this  example,  ^  of  the  sum 
of  9  and  3  is  4.  Adding  13  the  result  is,  as  the  example  requires,  17. 
Again  ^  the  difference  of  9  and  3  is  3.  Subtracting  1,  the  remainder 
is  2,  as  required.  ] 

4.  What  fraction  is  that,  whose  numerator  being  doubled, 
and  denominator  increased  by  7,  the  value  becomes  f; 
but  the  denominator  being  doubled,  and  the  numerator 
increased  by  2,  the  value  becomes  f  ?  Ans.,  f . 

SuG. — The  pupil  should  ask  himself,  "How  many  sets  of  con- 
ditions in  this  problem  ?"  "  What  are  they  ?"  "  How  many  unknown 
(required)  quantities  ?"  "  W^hat  are  they  ?"  There  must  always  he  as 
many  of  one  as  of  the  other.  The  unknown  (required)  quantities  here 
are  the  numerator  and  the  denominator  of  the  fraction.     K  these  are 

X 

called  respectively  x  and  y,  the  fraction  is  — .    Now,  by  the  first  set  of 

2a:  a:  -}-  2 

conditions, j — -  =  |,  and,  by  the  second  set  ' 


1/  4-  7        -         —  '       2y 

5.  What  fraction  is  that  which  becomes  f  when  its  nu- 
merator is  increased  by  6,  and  i  when  its  denominator 
is  diminished  by  2  ?  Ans.,  sV 

6.  If  1  be  added  to  the  numerator  of  a  certain  fraction, 
its  value  is  ^;  but  if  1  be  added  to  its  denominator,  its 
value  is  \.     What  is  the  fraction  ?     Verify. 

Ans.,  y4^ 


268  SIMPLE  EQUATIONS 

7.  There  is  a  certain  number,  to  the  sum  of  whose  digits 
if  you  add  7,  the  result  will  be  three  times  the  left 
hand  digit ;  and  if  from  the  number  itself  you  subtract 
18,  the  digits  will  have  changed  places.  AVhat  is  the 
number?     Verify.  Ans.,   53. 

SuG. — The  two  numbers  sought  are  the  two  digits.  Hence  let  y  =3. 
the  units  digit,  and  x  =  the  tens  digit.  The  number  tiien  is  lOx  -f-  y. 
(Just  as  when  6  is  the  units  digit  of  a  number  and  5  the  tens,  the  num- 
ber is  10  •  5  +  C.  Of  course  the  number  would  not  be  represented  by 
xy,  for  this  would  indicate  the  product  of  the  digits.  (See  Part  I.,  30, 
Second  Law,  Scholium  1st. )  The  first  conditions  give  2x  —  y  =  7, 
and  the  second  lOx  +  .V  —  18  =  lOy  -j-  x,  i.  e.  the  units  becomes  the 
tens  figure  and  t&©  tens  becomes  the  units. 

8.  A  certain  number  of  two  digits  contains  the  sum  of  its 
digits  four  times  and  their  product  twice.  What  is  the 
number  ?  Ans.,  36. 

9.  There  is  a  number  consisting  of  two  digits;  the  num- 
ber is  equal  to  3  times  the  sum  of  its  digits,  but  if  the 
number  be  multiplied  by  3,  the  product  equals  the 
square  of  the  sum  of  its  digits.  What  is  the  number  ? 
Verify. 

SuG. — The  equations  are  10a:;  -f  V  =  ■^{^  -\-  ?/\  and  3(10x  -}-  y)  = 
(x  -\-  yY:  Multiplying  the  1st  by  3,  we  have  3(10.c  -\-y)  =  9(x  -^  y). 
Equating  (placing  equal)  the  second  members,  9  x  -f-  y-  =  (^c  -f-  y'^. 
Dividing  by  a;  -{-  y,  9  =  a:  -f-  ?/•  This  equation  and  the  first  are  easily 
combined  and  solved. 

10.  .A  number  consisting  of  2   digits,  when  divided  by  4 
gives  a  certain  quotient  and  a  remainder  of  3;  when 
divided  by  9,  gives  another  quotient  and  a  remainder 
of  8.     Now,  the  value  of  the  digit  on  the  left  hand  is 
equal  to  the  quotient  which  was  obtained  when  the 
number  was  divided  by  9;  and  the  other  digit  is  equal 
to  yV  ^^  ^^^  quotient  obtained  when  the  number  was 
divided  by  4.     What  is  the  number  ?     Verify. 
SuG. — Letting  x  represent  the  tens  figure,  and  7/ the  units,  the  equa- 
tions are — '—- — -  =  lly  -]-\,  and  — '  =  x   -f-   f-      Th^  pupil 
should  give  the  explanation. 


WITH   TWO   UNKNOWN    QUANTITIES.  269 

11.  A  farmer  parting  with  his  stock,  sells  to  one  person  9 

horses  and  7  cows  for  300  dollars;  and  to  another,  at 
the  same  prices,  6  horses  and  13  cows  for  the  same 
sum.     What  was  the  price  of  each  ? 

Ans.,  $24  and  $12. 

12.  A  son  asked  his  father  how  old  he  was.  His  father 
answered  him  thus :  If  you  take  away  5  from  my  years, 
and  divide  the  remainder  by  8,  the  quotient  will  be  ^ 
of  your  age ;  but  if  you  add  2  to  your  age,  and  multi- 
ply the  whole  by  3,  and  then  subtract  7  from  the  pro- 
duct, you  will  have  the  number  of  the  years  of  my  age. 
What  was  the  age  of  the  father  and  son  ? 

Ans.,  53  and  18. 

13.  A  farmer  purchased  100  acres  of  land  for  $2450;  for  a 
part  of  the  land  he  paid  $20  an  acre,  and  for  the  other 
part  $30  an  acre.  How  many  acres  were  there  in  each 
part  ?     Verify. 

ScH. — Very  many  such  problems  can  be  solved  equally  weU  by 
means  of  one  or  of  two  unknown  quantities.  The  pupil  should  do  such 
in  both  ways. 

14.  At  a  certain  election  946  men  voted  for  two  candidates, 
and  the  successful  one  had  a  majority  of  558.  How 
many  votes  were  given  for  each  candidate  ?     Verify. 

15.  A  jockey  has  two  horses  and  two  saddles.  The  saddles 
are  worth  15  and  10  dollars,  respectively.  Now  if  the 
better  saddle  be  put  on  the  better  horse,  the  value  of 
the  better  horse  and  saddle  will  be  worth  ^  of  the 
other  horse  and  saddle.  But  if  the  better  saddle  be  put 
on  the  poorer  horse,  and  the  poorer  saddle  on  the  bet- 
ter horse,  the  value  of  the  better  horse  and  saddle  will 
be  worth  once  and  -^^  the  value  of  the  other.  Required 
the  worth  of  each  horse  ?       Besult,  65  and  50  dollars. 

16.  A  sum  of  money  was  divided  equally  among  a  certain 
number  of  persons  ;   had  there  been  four  more,  each 


270  SIMPLE    EQUATIONS 

would  have  received  one  dollar  less,  and  had  there  been 
four  fewer,  each  would  have  received  two  dollars  more 
than  he  did  :  required  the  number  of  persons,  and 
what  each  received  ?     Verify. 

SuG.    -= — -— -+ljand-= 2.      Hence  xy  4- 4x  =  xy  4- 

y     y+^  y     y-^  ^^  ^^ 

y-  -f-  4?/,  and  xy  —  4x  =  xy  —  2?/'  -f-  8?/,  or  4ic  =  ?/-  -f-  4?/,  and  —  2x  = 
—  ?/2  -j-  4i/.     Adding  2x  =  8y. 

17.  A  farmer  hired  a  laborer  for  ten  days,  and  agreed  to 
pay  him  $12  for  every  day  he  labored,  and  he  was  to 
forfeit  $8  for  every  day  he  was  absent.  He  received  at 
the  end  of  his  time  $40.  How  many  days  did  he  labor, 
and  how  many  days  was  he  absent  ?     Verify. 

18.  A-  boatman  can  row  down  stream  a  distance  of  20  miles, 
and  back  again,  in  10  hours,  the  current  being  uniform 
all  the  time  ;  and  he  finds  that  he  can  row  2  miles 
against  the  current  in  the  same  time  that  he  rows  3 
miles  with  it.  Required  the  time  in  going  and  return 
ing.  Result,  4  and  6  hours. 

SuG. — If  £c  and  ?/  are  the  times  of  rowing  down  and  up,  respectively, 
at  what  rate  does  he  row  down  ?  At  what  rate  up  ?  Twice  one  of 
these  rates  equals  3  times  the  other. 

19.  A  and  B  together  could  have  completed  a  piece  of  work 
in  15  days,  but  after  laboring  together  6  days,  A  was 
left  to  finish  it  alone,  which  he  did  in  30  days.  In  how 
many  days  could  each  have  performed  the  work  alone  ? 

An^.,  50,  and  21f  days. 

SuG. — If  a;  represent  the  number  of  days  A  would  require  to  do  it 
alone,  and  y  the  number  B  would  require,  how  much  would  each  do 
in  a  day?  How  much  both?  How  much  would  they  do  in  6  days? 
How  much  would  remain  to  be  done  by  A  alone  ?  How  much  would  A 
do  in  30  days  ?     In  resolving  these  equations  do  not  clear  of  fractions. 

20.  Two  pipes,  the  water  flowing  in  each  uniformly,  filled 
a  cistern  containing  330  gallons,  the  one  running  during 
5  hours,  and  the  other  during  4  ;  the  same  two  pipes, 
the  first  running   during  two  hours,  and  the  second 


WITH   TWO   UNKNOWN   QUANTITIES.  271 

three,  filled   another   cistern   containing   195  gallons. 
The  discharge  of  each  pipe  is  required.     Verify. 

21.  If  I  were  to  enlarge  my  field  by  making  it  5  rods  longer 
and  4  rods  wider,  its  area  would  be  increased  240 
square  rods  ;  but  if  I  were  to  make  its  length  4  rods 
less,  and  its  width  5  rods  less,  its  area  would  be  dimin- 
ished 210  square  rods.  Required  the  present  length, 
width,  and  area.     Verify. 

22.  A  farmer  sells  a  horses,  and  h  cows  for  $7?i ;    and  at  the 

same  prices  a^  horses,  and  h^  cows  for  %mi;  what  is  the 

price  of  each  ?     Apply  the  results  to  Ex.  11. 

^-      -,  J>\^n — ftm,        -  .o,m  —  an\ 

Ans.,  Of  a  horse  $ — ; —  ;  of  a  cow  $ — 7 r— . 

a6i  —  a^h  ajb  —  061 

[Note. — Observe  the  symmetry  of  such  results.  Thus,  in  these  nu- 
muerators  the  a  and  b  change  places  and  in  the  denominators  the  sub- 
scripts change  letters.] 

23.  A  man  bought  s  acres  of  land  for  $m.     For  a  part  he 

paid  %a  per  acre,  and  for  the  rest  %a^  per  acre.     How 

many  acres  in  each  part?     Deduce   from  the  general 

answer  obtained  in  this  case  the  particular  answers  to 

^     ,^  .        m  —  a,s        _  m  —  as 

Ex.  13.  Ans.,  and  acres. 

a — fli  fli  —  a 

24.  A  waterman  rows  a  given  distance  a  and  back  again  in 
h  hours,  and  finds  that  he  can  row  c  miles  with  the  cur- 
rent for  d  miles  against  it :  required  the  times  of  row- 
ing down  and  up  the  stream,  also  the  rate  of  the  cur- 
rent and  the  rate  of  rowing  ? 

rr,.        -.  hd        ,.  be  ,        - 

Ans.,  Time  down, ;  time  up,  — — ; ;    rate   of 

c-^-d  c-{-d 

a{cr-  —  d-^)         ,      .        .       a(e-hd)^ 
current,  -—z —  ;  rate  of  rowmg, 


2bed      '  *'     '2bcd 

Deduce  from  these  answers  those  of  Ex.  18. 


272  SIMPLE    EQUATIONS 

[Note.— Several  varieties  of  examples  usually  given  in  this  place  are 
reserved  for  their  proper  places  under  the  discussions  of  the  general 
principles  or  problems  of  which  they  are  special  examples.  Such  are, 
examples  involving  ratio,  proportion,  percentage,  aUigation,  etc.  ] 


SECTION  III 


Simple,  Simultaneous,  Independent  Equations  with 
more  than  Two  Unknown  Quantities. 

40*  JProh.  Having  given seiwral  simple,  simultayieoiis, in- 
dependent equations,  involving  as  many  unhioivn  qiianti- 
ties  as  there  are  equations,  to  find  the  values  of  tlie  unknoiun 
quantities. 

RULE.. — Combine  the  equations  two  and  two  by  either 

OF  THE  METHODS  OF  ELIMINATION,  ELIMINATING  BY  EACH  COMBINA- 
TION THE  SAME  UNKNOWTSf  QUANTITY,  THUS  PRODUCING  A  NEW  SET 
OF  EQUATIONS,  ONE  LESS  IN  NUMBER,  AND  CONTAINING  AT  LEAST  ONE 
LESS  UNKNO^TSf  QUANTITY.  CoMBINE  THIS  NEW  SET  TWO  AND  TWO 
IN  LIKE  MANNER,  ELIMINATING  ANOTHER  OF  THE  UNKNOWN  QUAN- 
TITIES. Repeat  the  process  until  a  single  equation  is  found 
with  but  one  unknown  quantity.  solve  this  equation  and 
then  substitute  the  value  of  this  unknown  quantity  in  one 
of  the  next  preceding  set  of  equations,  of  which  there 
will  be  but  two,  with  two  unknown  quantities,  and  there 
w^ll  result  an  equation  containing  only  one  and  that 
another  of  the  unknown  quantities,  the  value  of  which  can 
therefore  be  found  from  it.  substitute  the  two  values 
now  found  in  one  of  the  next  preceding  set,  and  find  the 
value  of  the  remaining  unknown  quantity  in  this  equation. 
Continue  this  process  till  all  the  unknown  quantities  are 
determined. 

ScH.l. — If  any  equation  of  any  set  does  not  contain  the  quantity 
you  are  seeking  to  eUminate  from  the  following  set,  this  equation 


WITH   MORE   THAN   TWO   UNKNOWN   QUANTITIES.      273 

cau  be  written  at  once  in  that  set  and  the  remaining  equations  com- 
bined, 

ScH.  2. — In  ehminating  any  unknown  quantity  from  a  particular  set 
of  equations,  any  one  of  the  equations  may  be  combined  with  each  of 
the  others,  and  the  new  set  thus  formed.  But  some  other  order  may 
be  preferable  as  giving  simpler  results, 

ScH.  3. — It  is  sometimes  better  to  find  the  values  of  all  the  unknown 
quantities  in  the  same  way  as  the  first  is  found,  rather  than  by  substi- 
tution. 

Dem.  1. — The  combinations  of  the  equations  give  true  equations  be- 
cause they  are  all  made  upon  the  methods  of  eUmination  akeady  de- 
monstrated, 

2,  That  the  number  of  equations  can  always  be  reduced  to  one  by 
this  process,  is  evident,  since,  if  we  have  n  equations  and  combine  any 
one  of  them  with  each  of  the  others,  there  will  be  ?i  —  1  new  equations. 
Combining  one  of  these  n  —  1  new  equations  with  all  the  rest  there 
will  result  n  —  2.  Hence  n  —  1  such  combinations  will  produce  a  sin- 
gle equation;  and  as  one  unknown  quantity,  at  least,  has  disappeared 
from  each  set  there  will  be  but  one  left.     q.  e.  d. 


EXAMPT,FS. 

1.  Given 

(10 

1X  —  2Z  +  du  =  17, 

(20 

^  +  4?/  —  2z  =  11, 

(30 

5y  —  Sx  —  2iL=    8, 

(40    - 

-  3w  ^2t  +  4ty=    9, 

(50 

Sz  +  8a  =  33,  to  find  the  value 

of  X,  y, 

z,  t,  and  u. 

MODEL   SOLUTION. 

OPEBATION. 

(1*) 

t^iy-2z=ll- 

(2,) 

2«  -f-  4?/  —  3u  =     9         2nd  set ,  from  which 

(3*) 

3z-f-8u=  33  rx  is  absent. 

(4,) 

35y  —  62  —  5u  =107 

(I3) 
(23) 
(33) 

Sz-{-8u=  33-] 
35r/  -  6s  -  5u  =  107  L      ^rd  set,  from  which 

4y  -  4.  4-  3u  =  13  J  ^  ^^^  ^  ^^^  ^^'^^^• 

(I4) 

3r  4-      8it  =       33  )      4th  set,  fi-om  which 

(2,) 

116z  —  125  a  =  —  27  fa;,  t,  and  y,  are  absent. 

274: 


SIMPLE   EQUATIONS 


(I5)     lS03u 


3909 


iu) 

3z  +  21  =  33 

.-.  z  =  3 

(33) 

47/  _  12  +  9  =  13 

•••  2/  =  4 

(12) 

i_f_  16  — 6  =  11 

.-.    f  =  l 

(3x) 

20  — 3a3  — 6=    8 

.-.  x  =  2 

Explanation. — I  notice  tliat  I  have  5  equations  with  5  unknown 
quantities.  From  these  I  wish  to  produce  a  new  set  of  4  equations 
from  which  one  at  least  of  the  unknown  quantities  shall  be  eliminated. 
I  observe  that  x  does  not  appear  in  (2i),  (4,),  and  (Sj),  hence  I  write 
these  as  three  of  the  2d  set  of  equations.  Then  eUminating  x  between 
(li)  and  (3i)  I  have  (42),  and  thus  obtain  the  2nd  set  of  4  equations 
containing  only  4  unknown  quantities.  (If  desirable  the  pupil  may  be 
asked  how  he  knows  that  (I2)  is  a  true  equation.) 

Again,  as  t  is  contained  in  a  less  number  of  this  set  of  equations  than 
any  one  of  the  other  unknown  quantities,  leUminate  it  next;  i.  e.,  I 
produce  a  3rd  set  which  does  not  contain  it.  As  (3,j)  and  (4j)  do  not 
contain  t,  I  transfer  them  at  once  to  the  3rd  set;  and  then  ehminating 
t  between  (Ig)  and  (23)  this  set  is  complete,  having  3  equations  with  3 
unknown  quantities. 

Now  eliminating  y  from  this  set  by  combining  (23)  and  (83),  and 
transferring  (I3),  I  have  the  4th  set  of  two  equations  with  only  2  un- 
known quantities.  Combining  these  two  so  as  to  eliminate  z  I  find 
M  =  3. 

Finally,  substituting  3  for  u  in  (I4),  I  find  z  =  3. 
Substituting  3  for  u  and  3  for  z  in  (33),  I  find  y  =:^  4,. 
Substituting  4  for  y  and  3  for  z  in  (I2),  I  find  t  ==  1. 
Substituting  the  values  oiy  and  u  in  (3j ),  I  find  x  ==  2. 


K 


2.  Given    -^  ^  -f  z  =  25,  V  Values, 

(y+  z  =  15.) 

/  8j;  —  4?/  =  24  —    z,\ 

3.  Given    -|  6^  +    y  ^     2;  +  84,  V      Values, 

[    a;  +  80  =  3?/  +  42.  j 

f  3u  +    a:  +  2?/ 


4    Given    -\ 


z  =  22,  ] 

4.x—    2/  +  3z  =  35,  !  . 

Y  Values,  -\ 


^u  +  3:r  —  2?/ 


19, 


L 


2u  -^  4y  -\-2z=  46.  J 


WITH  MORE  THAN  TWO  UNKNOWN  QUANTITIES.   275 


5.    Given 


X 

2 

+ 

i-^ 

2 

X 

a 

+ 

!+ 

5 

X 

4 

+ 

\- 

z 

=  124, 

=    94, 

= "  76. 


Results,  - 


48, 
120, 
240. 


SuG. — Such  examples  afford  opportunity  for  the  exercise  of  no  Httle 
ingenuity,  m  order  to  avoid  large  numbers  and  inconvenient  fractions. 
For  example,  this  may  be  solved  in  the  ordinary  way  by  clearing  of 
fractions,  etc. ,  but  the  following  is  far  more  elegant : 

Dividing  the  1st  by  3  and  the  2nd  by  2  and  subtracting,  we  have 

11  ) 
"3~ 


y_  A.^ 

72  "^  GO 
Subtracting  ^  the  1st  from  the  3rd 30"^  2r 


2nd  set. 


14 


Or, 


'dm  "*"  5-60       15 


360  ^24- 12 


Subtracting 

24-1^.^^- 

=  30    ^^   240-^' 

2  5 

1 

>h-l 

r.= 

-^  6.    Given    - 

- 

Values,  -< 

=  1,  and  z  =  240. 


a  +  6  —  c 
2 


6  +  c  —  a 


SuG. — Do  not  clear  of  fractions.  Having  found  the  value  of  one  un- 
known quantity,  do  not  get  the  others  by  substitution,  but  return  to 
the  original  equations  and  get  each  in  the  same  manner. 


—  7.  Given 


2 

X 

j3 


3       4 

y     z 

4      5      6. 


X 


y 


x=    6, 


Values,]    2/ =  12, 


276 


SIMPLE    EQUATIONS 


8.  Given  J 


9.  Given, 


y  -\-  a  :^  2x  -{-  2z,  [  Values,  ^  y 


9^+6y  +  4z  +  3u  +  2w 
+  1  =  0, 

16^  +  4i;  +  1  =  0, 

25^ — ^y  -\-  z-\-bv — u-\- 
1=0, 
X  -\-  1y  +  ^z  —  V —  2u 
+  1  =  0, 
4:X  —  2y-\-z  —  2v  +  u 
+  1  =  0.  j 


Values,  - 


zi  +  V  +  ^  +  ?/  =  10, 
If  +  u  +  ^'  +  2  =  11, 
10.  Given,  \  u  -\-  v  -^  y  +  z  =12,  \ 
u  +  ^  +  ?/  +  z  =  13,  I 
(;  +  :^  -f  2/  +  2  =  14.  J 

f  3.f/  —  1      62       j; 


12/ 

Values,  \  z 

I 


11.  Given, 


4  5       2 

5.r      42  5 

1 =  1/  -1- 


_i_  14 
T^  -'■5J 


T=y+& 


-    Fa/i^e. 


3^  +  1 


2^  ,  y 


+  ^  =  777  + 


'*■> 


14  '   6       21   '    3  J 
12,v  — 11^ 


11^  —  lOy 


12.  Given,  \  x-\-z  —  2y       z  —  y  —  1 


Values,  - 


3 
Zx 


2/  +  2  +  7. 


y  = 


a 

il' 

5a 
II' 

la 

ir 

169 
~924' 
220 
924' 
_89_ 
924' 
4^5 
924' 
113 
924* 

=  3, 
=  4, 
=  5, 
=  2, 
=  1. 

=  2, 

=  3, 

=  1. 

10, 

11, 
12. 


WITH    MOKE   THAN    TWO    UNKNOWN    QUANTITIES.       277 


13.  Given, 


'  Values, 


^=9, 


!/-7, 


z  =  3. 


14.  Given,  ■{  }j(x  -\-  z)  =  d  —  ij, 

i(^_z)  =  2^-7. 


Values,  -J  ?/  =  4, 
z  =3. 


APPLICATIONS. 

1.  The  sum  of  three  numbers  is  9.  The  sum  of  the  first, 
twice  the  second,  and  three  times  the  third  is  22.  The 
sum  of  the  first,  four  times  the  second,  and  nine  times 
the  third  is  58.     What  are  the  numbers  ? 

Ans.,  1,  3,  and  5. 

SuG. — How  many  unknown  quantities?  How  many  sets  of  condi- 
tions? What  are  they?  Express  the  first  in  an  equation, — the 
second, — the  third. 

The  form  of  explanation  is  the  same  as  in  the  last  section. 

2.  Five  persons,  A,  B,  C,  T>,  and  E  played  at  cards ;  after  A 
had  won  half  of  B's  money,  B  one-third  of  C's,  C  one- 
fourth  of  D's,  and  D  one-sixth  of  E's,  they  each  had 
$7.50.      How  much  had  each  to  begin  with? 

Ans.,  A,$2.75;  B  $9.50,  C  $8.25;  D,$8,  and  E,$9. 

3.  There  are  4  men.  A,  B,  C,  and  D,  the  value  of  whose 
estates  is  $14,000  ;  twice  A's,  three  times  B's,  half  of 
C's,  and  one-fifth  of  D's,  is  $16,000  ;  A's,  twice  B's, 
twice  C's,  and  two-fifths  of  D's,  is  $18,000  ;  and  half  of 


278  SIMPLE    EQUATIONS 

A's,  with  one- third  of  B's,  one-fourth  of  C's,  and  one- 
fifth  of  D's,  is  $4,000.  Kequired  the  property  of  each, 
Ans.,  A's,$2,000;  B's,$3,000;  C's,$4,000;  r>'s^$5,000. 

4.  A  number  is  expressed  by  three  figures  ;  the  sum  of 
these  is  11  ;  the  figure  in  the  place  of  units  is  double 
that  in  the  place  of  hundreds,  and  when  297  is  added 
to  this  number,  the  sum  obtained  is  expressed  by  the 
figures  of  this  number  reversed.     What  is  the  number? 

Ans.,  326. 

SuG. — Letting  x  represent  the  hundreds  figure,  y  the  tens,  and  z  the 
Units,  the  number  is  represented  by  lOOx  -f-  lOy  -f-  z.  The  number 
with  the  digits  reversed  is  lOOz  -f  lOy  -f  a;. 

5.  A  man  worked  for  a  person  ton  days,  having  his  wife 
with  him  8  days,  and  his  son  6  daj^s,  and  he  received 
$10.30  as  compensation  for  all  three  ;  at  another  time 
he  wrought  12  days,  his  wife  10  days,  and  son  4  days, 
and  he  received  $13.20  ;  at  another  time  he  wrought 
15  days,  his  wife  10  days,  and  his  son  12  days,  at  the 
same  rates  as  before,  and  he  received  $13.85.  What 
were  the  daily  wages  of  each  ? 

Ans.,  The  husband  75  cts.;  wife, 50  cts.  The  son,  20  cts. 
expense  per  day., 

SuG. — Solving  the  equations  which  express  the  conditions,  the  value 
of  the  quantity  representing  the  son's  wages  is  found  to  be  negative. 
But  as  negative  quantities  are  such  as  are  opposed  in  their  nature  to 
those  called  positive  in  the  same  problem,  the  son  produced  the  oppo- 
site effect  from  wages. 

6.  Three  masons,  A,  B,  C,  are  to  build  a  wall.  A  and  B, 
jointly,  can  build  the  wall  in  12  days  ;  B  and  C  can 
accomplish  it  in  20  days,  and  A  and  C  in  15  days. 
How  many  days  would  each  require  to  build  the  wall, 
and  in  what  time  will  they  finish  it,  if  all  three  work 

'  together  ? 

Ans.,  A  requires  20  days;  B,30;and  C,60  ;  and 
all  three  require  10  days. 


WITH   MORE   THAN   TWO   UNKNOWN   QUANTITIES.      279 

7.  Three  laborers  are  employed  on  a  certain  work.  A 
and  B,  jointly,  can  complete  the  work  in  a  days  ;  A 
and  C  require  h  days,  B  and  C  require  c  days.  What 
time  does  each  one,  Avorking  alone,  require  to  accom- 
plish the  work,  on  the  condition  that  each  one,  undet 
all  circumstances,  does  the  same  quantity  of  work? 
And  in  what  time  would  they  finish  it,  if  they  all  three 
worked  together  ? 

Ans..  A  requires  , days,  B  r — ■ — 7 

he  -\-  ac  —  ab  be  -\-  ab  —  ac 

days,  and  C  -- ; r~  days. 

-^  '  ab  ^  ac  —be      "^ 

Jointly,  they  require  ^^  ^  ^^  ^  ^^  days. 

Deduce  from  these  results  those  of  the  preceding  ex- 
ample. 

8.  If  A  and  B  together  can  perform  a  piece  of  work  in  8 
days,  A  and  C  together  in  9  days,  and  B  and  C  to- 
gether in  10  days,  in  how  many  days  can  each  alone 
perform  the  same  -svork  ? 

Ans.,  A  in  14f  *  days,  B  in  11 1^  days,  and  C  in 
233V  days. 

9.  A  gentleman  left  a  sum  of  money  to  be  diyided  among 
his  four  sons,  so  that  the  share  of  the  oldest  was  -^  of 
the  sum  of  the  shares  of  the  other  three,  the  share  of 
the  second  ^  of  the  sum  of  the  other  three,  and  the 
share  of  the  thu'd  \  of  the  sum  of  the  other  three;  and 
it  was  found  that  the  share  of  the  oldest  exceeded  that 
of  the  youngest  by  $14,  What  was  the  whole  sum, 
and  what  was  the  share  of  each  person  ? 

Ans.,  Whole  sum,  $120  ;  oldest  son's  share.,  $40  ; 
second  son's,  $30;  third  son's,  $24  ;  youngest 
son's,  $26. 


280  SIMPLE   EQUATIONS. 

Synopsis. 

r  Algebra. 
Equation.     Members. 
Defs.  \  Independent  equations.     Simultaneous. 

I  Transposition.     Elimination. 

[  Statement.     Solution. 
rr-    J      f     {  Simple,       )  With  one  unknown  quantity. 
Kinds  of     jQ^jadratic,  V 

j^quauon^.    j  higher.       )  With  more  than  one  nnk'n  quant. 
Axioms. 

r  1.  Clearing  of  frac's.  Kule.     Dem.    111. 

m        /.         ^-  2.  Transposition.     Rule.     Dem.     111. 

Transformations.       ^   -^^.^.^^  ^^^,^^^ 

[  4.  Dividing  by  coeff.  of  unk'n  quant. 

Proh.  1.   To  solve  simple  equations.       )  „  p         .„„^p^tioTm 
Rule.     Dem.  \  '  ^^^^-  suggestions. 

Proh.  2.  To  free  an  equation  of  radicals  (  ^  -r.  i.- 

Rule.     Dem.  [  6  Prac.  suggestions. 

!^   f  Number  of  methods.     Reason  for  several. 

H    [  Prob.  1.  By  comparison.     Rule.     Dem. 

<  J  Prob.  2.   By  substitution.     Rule.     Dem. 

M   I  Py^ob.  3.  By  addition  and  subtraction.     Rule.     Dem. 

M      Prob.  4.  With  several  unknown  quan-  )  q^,    i    o    q 

i  titles.     Rule.     Dem.  \  ^^^-  ■^'  ^^  ^' 


Test  Questions. — Upon  what  principle  is  an  equation  cleared  of 
fractious  ?  How  is  it  done  ?  Upon  what  principle  is  elimination  by 
addition  and  subtraction  performed  ?  What  comparison  ?  Substitu- 
tion ?  Give  the  seven  Practical  Suggestions  upon  solving  Simple 
Equations.  The  six  upon  freeing  of  Radicals.  Give  the  reason  for 
changing  the  signs  of  the  terms  of  a  fraction  having  a  poljmomial  nu- 
merator, preceded  by  a  minus  sign,  when  clearing  of  fractions.  What 
is  the  general  method  of  procedure  in  stating  a  problem  ?  Does  the 
statement  involve  a  knowledge  of  anything  but  algebra  ?  Illustrate. 
Upon  what  principle  may  all  the  signs  of  an  equation  be  changed  ? 
(This  may  be  explained  in  at  least  four  different  ways.)  Having  given 
the  sum  and  difference  of  two  quantities,  how  are  the  quantities  found  ? 
Prove  it. 


RATIO,  PROPORTION,  AND  PROGRESSION.      281 

CHAPTEE  11. 

RATIO,  PROPORTION  AND    PROGRESSION, 


SUCTION  I. 
Eatio. 

47.  Mcitio  is  the  relative  magnitude  of  one  quantity 
as  compared  with  another  of  the  same  kind,  and  is  ex- 
pressed by  the  quotient  arising  from  dividing  the  first  by 
the  second.  The  first  quantity  named  is  called  the  Antece- 
dent, and  the  second  the  Consequent.  Taken  together  they 
are  called  the  Tei^ms  of  the  ratio,  or  a  Couplet. 

48,  The  Sign  of  ratio  is  the  colon,  :  ,  the  common 
sign  of  division,  -i-  or  the  fractional  form  of  indicating 
division. 

Q 

III. — The  ratio  of  8  to  4  is  expressed  8  :  4,  8  -^  4,  or  _,  any  one  of 

4 

>  whicli  may  be  read  "  8  is  to  4,"  or,  "  ratio  of  8  to  4."  The  aniecedeni 
is  8,  and  the  consequent  4.  The  sign  :  is  an  exact  equivalent  for  -j-  , 
and  by  many  writers,  especially  the  Germans,  the  former  is  used  ex- 
clusively. The  sign  :  is,  probably,  a  mere  modification  of  -f-,  made 
by  dropping  the  horizontal  line,  as  unnecessary.  Possibly  the  sign  ~ 
finds  its  analogy  to  the  fractional  form  of  expressing  division,  by  con- 
sidering the  upper  dot  as  symbohzing  a  dividend,  and  the  lower  a 
divisor. 

40,  CoR. — A  ratio  being  merely  a  fraction,  or  an  unexe- 
cuted problem  in  Division,  of  which  the  antecedent  is  the  nu- 
merator, or  dividend,  and  the  consequent  the  denominator,  or 
divisor,  any  changes  made  upon  the  terms  of  a  ratio  produce 


282  RATIO,   PROPORTION,  AND   PROGRESSION. 

the  same  effect  upon  its  value,  as  the  like  changes  do  upon  the 
value  of  a  fraction,  lohen  made  upon  its  corresponding  terms. 
The  principal  of  these  are, 

1st.  If  both  terms  are  multiplied,  or  both  divided  by  the  same 
number,  the  value  of  the  ratio  is  not  changed.  Thus,  the  ra- 
tio 16   :   8  is  2.     So,  also,  4  x  16  :  4  x  8,  or  64  :  32  is  2 ; 

16      8 
or  —  :  -,    i.  e.,  8  :  4  is  2. 

Ji        A 

2nd.  A  ratio  is  multiplied  by  multiplying  the  antecedent 
(i.  e.,  the  numerator  or  dividend),  or  by  dividing  the  con- 
sequent (i,  e.,  the  denominator,  or  divisor).  Thus,  32  :  8  is 
4,  but  2  X  32  :  8,  i.  e.,  64  :  8  is  2  x  4,  or  8.     So,  also,  32  : 

o 

-^,    i.  e.,  32  :  4,  is  2  X  4,  or  8. 

3rd.  A  ratio  is  divided  by  dividing  the  antecedent,  (i.  e.,  the 

numerator,   or  dividend),  or  by  multiplying  the  consequent, 

({.  e.,  the  denominator,  or  divisor).     Thus  24   :    6  is  4,  but 

24  4 

—  :  6,  i.  e.,  12  :  6  is  -,  or  2.  So,  also,  24  :  2  x  6,  i.  e., 
Z  A 

24  :    12  is  -,    or  2. 
A 

50.  A  Direct  Ratio  is  the  quotient  of  the  antece- 
dent divided  by  the  consequent,  as  explained  above.  {4t ,) 
An  Indirect  or  Meciprocal  Ratio  is  the  quotient 
of  the  consequent  divided  by  the  antecedent,  i.  e.,  the  re- 
ciprocal of  the  direct  ratio.  Thus,  the  direct  riitio  of  6  to 
3  is  2,  but  the  inverse  ratio  is  f  or  ^.  When  the  word  ra- 
tio is  used  without  qualification  it  means  direct  ratio.  The  in^ 
verse  or  reciprocal,  it  will  be  seen,  is  the  ratio  of  the  reciprocals. 
Thus  the  inverse  ratio  of  8  to  4  is  the  ratio  of  \  to  \,  or  ^. 

51,  A  ratio  is  always  written  as  a  direct  ratio.  Thus,  if 
the  inverse  ratio  of  6  to  2  is  required,  we  write  2  :  6.     The 

inverse  ratio  of  aiobi^b  -.  a,  or  -  :  -,  the  latter  beincr  ex- 

a    b 

pressed  as  the  direct  ratio  of  the  reciprocals. 


RATIO.  283 

52.  A  ratio  of  Greater  Inequality  is  a  ratio  which 
is  greater  than  unity,  as  4  :  3.  A  ratio  of  Less  Iiie^ 
quality  is  a  ratio  which  is  less  than  unity,    as  3  :  4. 

53.  A  Compound  Ratio  is  the  product  of  the  cor- 
responding terms  of  several  simple  ratios.  Thus,  the  com- 
pound ratio  a  :  h,  c  \  d,  m  :  n,  \?>  acm  :  bdn.  This  term  cor- 
responds to  compound  fraction.  A  compound  ratio  is  the 
same  in  effect  as  a  compound  fi'action. 

54.  A  Duplicate  Ratio  is  the  ratio  of  the  squares,  a 
trijMcatef  of  the  cubes,  a  suhduplicate^  of  the  square 
roots,  and  a  suhtriplicate,  of  the  cube  roots  of  two  num- 
bers.    Thus,  a'  :  b\  a^  :  b^,  Va  :  Vb,  and  "^a  :   '^b. 

EXAMPLES. 

1.  What  is  the  ratio  of  3am3  to  Qam"  ? 

Model  Solution. — Since  the  ratio  of  two  quantities  is  the  quotient 
of  the  antecedent  divided  by  the  consequent,  the  ratio  Zarn?  :  Gam'^  is 


3am^        1 

2. 

What  is 

the  inverse 

ratio  of 

a  — 

-  6  to  a2 

—  6-^? 

Ans.,  a 

+  6. 

3, 

What  is 

the  ratio  of 

2      1 

Of 

5  , 

6  ' 

■   3' 

a- 

a  —  x'> 

Of*^-='"to 

2cm  ^ 

Of*'' 

-y-  . 

X  — 

-y 

db^y       'db'^y-  x^ -\- y^      x- — xy-\-y^ 

Answers  to  three,  1,  1\,  and  1^. 

4.  What  is  the  triplicate  ratio  of  6  to  2.  Ans.,  27. 

5.  What  is  the  subduplicate  ratio  of  64  to  16  ?       Ans.,  2. 

6.  What  is  the  compound   ratio  of  3  to  4,  8  to  9,  2  to  G 

4 
and  4  to  2  ?  Ans.,  4  :  9,  or  -. 

y 

7.  Reduce  360  :  315  to  its  lowest  terms. 

SuG.--This  is  the  same  as  reducing  a  fraction  to  its  lowest  terms. 
The  result  is  8  :  7. 


284  KATIO,    PEOPORTION,  AND   PROGRESSION. 

8.  Eeduce  1595  :  667,  and  a^  +  2a"x  :  a^  to  their   lowest 
terms.  Result  of  the  last,  a  -\-  2x  :  1 . 

9.  Which  is  the  greater,  16  :  15,  or  17  to  14? 

SuG.  To  compare  two  fractions,  reduce  them  to  a  common  denom- 
inator. On  the  same  principle  these  ratios  become  224  :  210,  and 
255  :  210. 

10.  Which  is  greater,  a^  —  b^  :  a  —  6,  or  a^  -j-  2ah  -\-  b^  :  . 

a  +  bl 

11.  Which  is  least  of  the  ratios  20  :  17,  22  :  18,  and  25  :  23? 
Which  is  greatest,  8  :  7,  6  :  5,  or  10  :  9  ? 

12.  Which  is  greater,  a  +  2  :  ^a  +  4,  or  a  +  4  :  ^a  +  5  ? 

Alls.,  a  +  4:^a  +  5>>a  +  2  :ia  +  4. 

13.  What  is  the  compound  ratio  of  15  :  12,  6  :  7,  and 
9:4?  Ans.,  135  :  56. 

14.  Compound  the  ratios  a2 — ^2  •  a'i^a-^-x  :  b,  and  6  :  a — x. 

Result,  The  duplicate  ratio  of  a  -\-  x  io  a. 

15.  Show  that  the  compound  ratio  oi  x  -\-  y  \  a,  x  —  y  '.  b, 

and  0  :  IS  1. 

a 

16.  Is  the  compound  ratio  of  3^  +  2  :  6a  + 1,  and  2a  +  3  : 
a  +  2,  a  ratio  of  greater  or  of  less  inequality,  if  a  is 
—  f  ?    If  a  is  2,?     If  a  is  — 2? 

Ans.,  —  Zero.     Greater.     Infinity. 

17.  Compound  the  following  :  7  I  5,  the  duplicate  of  4  :  9, 
and  the  triplicate  of  3  :  2.  Result,  14  :  15. 

2 

Suggestion.     -  X  ^'4  X  ^-^,  --  14  :  15. 
5       0.0       t-t-t 
3 

18.  Compound  the  sub-duplicate  of  x'^  :  y^,  and  the  triph- 
cate  of  '^J7   :    "^y.  Result,  x^  :  yK 


PROPORTION.  285 

19.  Compound  the  inverse  ratio  of  vx  +  \/y  to   :c   —   y, 
and  the  direct  ratio  x  +  2V xy  -\-    y   :    Vx  +  Vy. 

Result,  X  —  y. 

20.  Which  is  the  greater,  the  inverse  subtriphcate  ratio  of 
8  to  64j  or  the  direct  dupUcate  ratio  of  2  to  3  ? 


SECTION   II. 
Proportion. 

do,  I*t*opoVtion  is  an  equahty  of  ratios,  the  terms 
of  the  ratios  bemg  expressed.  The  equahty  is  indicated  by 
the  ordinary  sign  of  equahty,  ^,  or  by  the  double  colon  :  :  . 
Thus,  8  :  4  =  6  :  3,  or  8  :  4   ;  ;  6  :  3,  or  8^  4  =  6  -^-  3, 

or  -  =  -;    all  mean  precisely  the  same  thing.     A  propor- 
tion is  usually  read  thus  :  "  as  8  is  to  4  so  is  6  to  3." 

ScH. — The  pupil  should  practice  writing  a  proportion  in  the  form 

a       c 

—  =:  — »    still  reading  it  "  a  is  to  6  as  c  is  to  d."     One  form  should  be  as 

familiar  as  the  other.     He  must  accustom  himself  to  the  thought  that 

a  :  h  \:  c  '.d  means  -r  =  -j  and  nothing  more.     It  will  be  seen  that  the 

o  <? 

language  "  8  is  to  4  as  6  is  to  3,"  means  simply  that  —  =  - ,    for  it  is 

4        o 

an  abbreviated  form  for  saying  that  "  the  relation  which  8  bears  to  4  is 

the  same  as  (is  equal  to)  that  which  6  bears  to  2  ;"   that  is,  8  is  as 

many  times  4  as  6  is  times  3,  or  -  =  -• 

4         6 

o6,  TJie  JExtr ernes  (outside  terms)  of  a  proportion 
are  the  first  and  fourth  terms.  The  JMecuiS  (middle  terms) 
are  the  second  and  third  terms.  Thus,  in  a  :  b  =  c  :  d, 
a  and  d  are  the  extremeSj  and  h  and  c  are  the  means. 


286  RATIO,    PROPORTION,  AND   PROGRESSION. 

S7»  A  Meet 71  Proportional  between  two  quanti- 
ties is  a  quantit}^  to  which  either  of  the  other  two  bears  the 
same  ratio  that  the  mean  does  to  the  other  of  the  two. 
Thus,  if  m  is  a  mean  proportional  between  a  and  6,  a  bears 
the  same  ratio  to  m  that  m  does  to  b;  i.  e.,  a   :  m  :  :  m  :  d. 

S8.  A  Third  l^roportional  to  two  quantities  is 
such  a  quantity  that  the  first  is  to  the  second  as  the  second 
is  to  this  third  (proportional).  Thus,  in  the  last  proportion, 
6  is  a  third  proportional  to  a  and  m.  So,  also,  a  is  a  third 
proportional  to  6  and  ra. 

ScH. — The  pupil  should  notice  carefully  the  language  used  in  the 
last  two  definitions.  We  do  not  say  "  a  mean  proportional  to,"  but  "a 
mean  jaroportional  between"  two  others.  So,  again,  we  say  "a  third 
proportional  to  two  others."  Moreover,  it  is  necessary  that  the  two 
others  be  taken  in  the  order  named  in  the  statement.  Thus,  if  ?/  is  a 
third  proportional  to  m  and  n,  m  :  n  :  :  n  :  y.  But,  if  y  is  a  third  pro- 
portional to  n  and  m,  n  :  m  :  :  m  :  y.  Notice  carefully  the  difference 
between  the  two  statements. 

SO,  A  proportion  is  taken  by  Tfiversion  when  the 
terms  of  each  ratio  are  written  in  inverse  order.  Thus,  if 
a  :  b  :  :  c  :  d,  hj  inversion  we  have  b  \  a  '.'.  d  '.  c.  It  is  to 
be  observed  that  in  inversion  the  means  are  made  extremes, 
and  the  extremes  means. 

00,  A  proportion  is  taken  by  Alternation  when  the 
means  are  made  to  change  places,  or  the  extremes.  Thus 
a  :  b  :  :  c  :  d  becomes  by  alternation  either  a  :  c  '. :  b  :  d, 
or  d  :  b  ::  c  :  a.  The  appositeness  of  the  term  alternation 
(taking  every  other  one)  is  seen  from  the  fact  that  the  new 
order  is  obtained  by  taking  the  terms  alternately  ;  that  is 
1st  and  3rd,  2d  and- 4th;   or  4tli  and  2nd,  3rd  and  1st. 

01,  A  proportion  is  taken  by  Composition  when 
the  sum  of  the  terms  of  each  ratio  is  compared  with  either 
term  of  that  ratio,  the  same  order  being  observed  in  both 
ratios;  or  when  the  sum  of  the  antecedents  and  the  sum  of 


PROPORTION.  287 

the  consequents  are  compared  with  either  antecedent  and 
its  consequent.  Thus,  if  a  :  b  ::  c  :  d,  by  composition 
we  have  a  +  b  :  a  ::  c  +  d  :  c,  or  a  +  b  :  b  ::  c  +  d  :  d,  or 
a  +  c  :  b  +  d  ::  a  :  b,  or  a  +  c  :  b  +  d  wed. 

02,  If  the  difference  instead  of  the  sum  be  taken  in  the 
last  definition,  the  proportion  is  taken  by  Division, 

03,  Four  quantities  are  Inversely  or  Reciprocally  Propor- 
tional when  the  1st  is  to  the  2nd  as  the  4th  is  to  the  3rd, 
or  as  the  reciprocal  of  the  3rd  is  to  the  reciprocal  of  the 
4th.  Thus,  if  a  and  h  are  to  each  other  inversely,  or  recipro- 
cally, as  m  and  n,  a  ;  6  :  :  n  :  m,  or  what  is  the  same  thing, 

.  ^     .    1        1 

a  ,  0  .:  —  :  — 

m       n 

04,  A  Continued  I^roportion  is  a  succession  of 
equal  ratios,  in  which  each  consequent  is  the  antecedent  of 
the  next  ratio.  Thus  ii  a  :  h  '.:  h  '.  c  :\  c  :  d  :\  d  \  eyVfQ 
have  a  continued  proportion. 


05.  Prop.  1.  In  a  proportion  the  product  of  the  ex- 
tremes equals  the  product  of  the  means. 

Dem.  —If  a  :h  :  :  c  :  d  then  ad  =  be.    For  a  :h  :  :  c  :  d  is  the  same 
a       c 
as  —  =  -yj    which  cleared  of  fractions  becomes  ad  =  he.     o.  e.  d. 
0        a 

00.  Cor.  1. — The  square  of  a  mean  proportional  equals  the 
product  of  its  extremes,  and  hence  a  mean  proportional  itself 
equals  the  square  root  of  the  product  of  its  extremes.  For,  if 
a  '.  m  :\  m  :  d,hj  the  proposition  m'^  =  ad.  Whence  ex- 
tracting the  square  root  of  both  members,  m  =  \/a^. 

07.  Cor.  2. — Either  extreme  of  a  proportioyi  equals  the 
product  of  tJie  means  divided  by  the  other  extreme;  and,  in 
like  manner,  either  mean  equals  the  product  of  the  extremes 


288 


RATIO,   PROPORTION,  AND   PROGRESSION. 


divided    by   the   other   mean.      For,    ii    a    :    b    :  :    c 

be  be    .        ad  ^  ad 

ad=bc.  .' .  d  =  — ,     a  =—-,  b  = — ,    and  c  =  -—. 
a  a  c  b 


d, 


08,  J*r02)»  2,  If  the  product  of  two  quantities  equals  the 
product  of  two  others,  the  two  former  maybe  made  the  extremes, 
or  the  means  of  a  proportion,  and  the  two  latter  the  other  terms. 


e.,  m  :x  ::n  :  y.    In  like  manner  dividing  by  mn  we  have 


Dem. — Suppose  my  =  nx.     Dividing  both  members  by  xy,  we  have 
m       n 
X   ~  y' 

n        in 

Let  the  pupil  determine  how  each  of  the  following  forms 
may  be  deduced  from  the  relation  my  =  nx. 


X.  e. 


n 


m. 


1. 

m 

:  X  : 

n 

2. 

m 

'.  n   '. 

X 

3. 

y 

n  : 

X 

4. 

X 

''  y  ' 

m 

5. 

y 

X  : 

n 

6. 

X  : 

m  : : 

y 

7. 

n  : 

m  : 

y 

8. 

n 

y  ' 

m 

y- 
y- 

m. 

n. 


m. 
n. 

X. 
X. 


Given  above. 
.  By  what  do  you  divide  ? 
,  Given  above. 
Dividing  mx  by  each  mem- 
ber we  have  -  = — 
y       ^ 
. .  By  what  do  you  divide  ? 
. .  How  obtained  ? 


TRANSFORMATIONS  OF  A  PROPORTION. 

09,  JProj)*  3.    Prop.  1,  together  with  the  two  principles 
that  such  changes  in  the  terms  of  a  proportion  may  be  made,  as, 

1.  Do  not  change  the  values  of  the  ratios, 

2.  Change  both  ratios  alike, 

are  sufficient  to  determine  in  all  cases  what  transformations  are 
possible  without  destroying  the  proportion. 

That  these  two  principles  are  correct  is  evident  from  the  nature  of 
a  proportion,  as  an  equality  of  ratios. 


PROPORTION.  289 

EXAMPLES. 
MULTIPLES. 

1.  If  a  :  6  w  X  \  y,  prove  that  ma  '.  mb  :  :  x  '.  y. 

Solution. — This  change  does  not  alter  the  value  of  the  first  ratio, 
and  hence  the  equality  of  ratios  remains. 

2  If  a  :  b  :  :  X  :  y  is  ma  :  mb  :  :  iix  :  ny  ?  Why  ?  Is 
the  value  of  either  ratio  changed  ?     Why  ? 

3.  11  a  \  b  '.'.  X  '.  y  is  ma  '.  b  '.'.  mx  :  y?  Is  the  value  of 
either  ratio  changed?     How  ? 

Ans.,  This  change  does  not  destroy  the  propor- 
tion, because  it  multiplies  both  ratios  by 
the  same  quantity. 

a  X 

4.  If  a  :  Z)  :  :  37  :  1/,  is  a  :  mb  \\  X  \  my  ?  or  —  \  b   '.'.  — 

:  1/  ?  or  —  '.  b  '.:  X  \  my  ? 

Answer  to  the  last.     Yes.     Both  ratios  are  divided  by 
m,  and  hence  the  equality  is  not  destroyed. 

5.  If  the  first  term  of  a  proportion  is  multiplied  by  any 
number,  in  what  four  ways  may  it  be  compensated  so 
as  not  to  destroy  the  proportion  ?  Does  multiplying 
the  third  term  by  the  same  number  compensate  ? 
Why?  Does  dividing  the  4th  term?  Why?  Does 
dividing  the  second  term  ?     Why  ? 

p.  If  the  third  term  of  a  j)roportion  is  divided  by  any 
number,  in  what  ways  can  the  change  be  compensated 
so  as  not  to  destroy  the  proportion  ?  Give  the  reason 
in  each  case. 


CHANGES  IN  THE  ORDER  OF  TERMS. 

7.  If  a  :  b  ::  X  :  y  is  a  :  x  \  :  b  :  ?/  ?     How  are  the  ratios 
changed  ? 


290  RATIO,    PROPORTION,  AND    PROGRESSION. 

Solution.  — To  ascertain  whether  the  proportion  a  :  x  :  :  b  :  y  is 

true,  I  must  see  if  the  ratios  are  equal,  that  is  if  —  =  —       Now,    re- 

X        y 

ducing  to   a  common  denominator,  these  ratios  become  —  and  -^. 

■xy  xy 

But,  from  a  :  6  :  :  x  :  y,  I  know  that  ay  =  hx     (03) .    .  • .    -  =  -,  i.  e. 

X        y 

the  proportion  a  :  x  :  :b  :  y  is  a  correct  deduction  from  a  :  b  :  :  x  :  y. 
To  ascertain  how  —  is  changed  so  as  to  become  -,  I  divide  the  latter  by 

the  former.     Thus  — '—  -  =  — ;  hence  —  has  been  multiphed  by   -. 

X     '   b         X  b  ^  -^    X 

But  —  X  —  =  — .    •  ■•  Both  ratios  a  :b  :  :  x  :  y  are  multiplied  by  —  in 
y       X       y  ^  ^  "^  X 

order  to  produce  a  :  x  :  :  b  :  y,  and  the  equality  is  not  destroyed  {60). 

[Note. — This  example  is  virtually  solved  in  the  demonstration  of 
Prop.  2  ;  but  the  purpose  here  is  to  exhibit  a  general  method  of  pro- 
cedure applicable  to  all  cases.] 

8.  If  m  \  X  '.'.  n  :  y,  is  n  :  m  :  :  y  :  x?      How   are   the 
ratios  changed? 

Ans.,  Yes.     The  first,  — ,  is  multiplied  by  — ,  and  the 

second,-,  is  multiplied  by^,  and  —  ^=  -^— , 
since  nx  =  my. 

9.  If  four  quantities  are  in  proportion  are  they  in  propor- 
tion by  inversion  ?     How  are  the  ratios  changed  ? 

Solution. — If  a  :  6  : :  c  :  d,  I  am  to  ascertain  whether  6  :  a  :  :  cZ  :  c, 

i.  e.,  whether  —  =— .     These  fractions  reduced  to  a    C.    D.   are  — 
a       c  ac 

and  - — .      But    from   the   given   proportion   I   know  that   ad  =  be. 
ac 

.  • .     —  =  -,  and  the  proportion  b  :  a  :  :  d  :  c  is  a  correct  deduction 
a         c 

from  a  :  b  : :  c  :  d.     Also,  as  another  method,  I  see  that  the  reciprocals 

of  the  ratios  of  the  tirst  proportion  are  taken  for  the  second  ;  and  two 

quantities  being  equal  their  reciprocals  are  equal. 

10.  If  3a'b  :  263  :  :  Omx  :  10m'X%  is  2  :  a-^  :  :  5mx  ;  b-^  ? 


PROPOETION.  291 

Suggestions. — We  are  to  ascertain  whether  — -  =  — --,  i.  e.,  if 

a-  t>^ 

. — 1  =  .     From  the  given  proportion,  by  Prop.  1,  ilh^mx  = 

30a-6m"a;-  or  2&2  =  Sa-ma;.     Therefore  the  answer  is,    Yes. 

11.  If  lax^  :  \hy  ::  a^x  :  ¥ij,  show  that  h  :  ^  \:  a^  '.  ^x. 


COMPOSITION      AND      DIVISION. 

12.  If  (2  :  6  :  :  7?i  :  ??,  show  that  a  -\-  h  \  a  '.  :  m  -{-  n  :  m. 

Solution. — I  am  to  prove  that    =  ,   knowing  that 

a  m 

a  :b  :  :  m  :n.     The  two  ratios  to  be  tested,  when  reduced  to  a  C.  D. 

am  -{-bm       ^  am  -\-  nn      .  .  ,  ,    ,  ,     .         .         ,, 

are and ,  which  are  seen  to  be  equal,  since  from  the 

am  am 

given  proportion  bm  =  an. 

13.  If  a  :  6  :  :  J7  :  ?/,  show  that  a  —  h  :  b  '.'.  x  —  y  '  y- 

14.  li  m  :  n  :  :  X  :  y,  show  that  m  -{-  n  :  m  —  n  ;:  x  -\-  y 
:  x  —  y. 

15.  If  ^a  —  X  :  ^a  -{-  X  w  h  ■ —  y  :  b  -{-  y,  show  that  2x  :  y 
'.'.  a  :  b. 

2a;  a 

Solution.— I  am  to  compare  — '-  and  -,  which  reduced  to  a  C.  D. 
y         b 

are  —  and  r^^— .     But  from  the  given  proportion  I  have,  by  Prop.   1, 

^ab  -{-bx  —  ^ay  —  xy  =  ^ab  —  bx  -f-  k^iy  —  xy,  which  reduced  gives 

2x        a 
2bx  =  ay.     .  • .     -^  =  -,  which  was  to  be  proved. 

16.  If  a  :  6  :  :  ^  :  y,  does  it  follow  that  a  —  y  :  b  —  x 
::  a  :  X?  Ans.,  No. 

17.  If  four  quantities  are  in  proportion,  does  it  follow 
that  they  are  in  proportion  by  composition,  or  by 
division,  separately,  or  by  both  at  once?  That  is  if 
a  \  b  '.'.  m  :  n,  is  a  +  ^  \  a  ::  m  -\-  n  \  m,  or  a  —  b 
:  a  ::  m  —  n  :  m,  or  a  +  6  :  a  —  b  ;  :  m-{-  n  :  m  —  n? 

Ans.,  Yes. 


292  RATIO,   PROPORTION,   AND   PROGRESSION. 

18.  li  a  :  b  ::  X  :  ^j  isa^  :  b^  ::  x-^  \  y-l    Is  a"  :  6"  : : 
X* :  2/"?   Is  v/a  :  v/^  : :   v/^j  :  v/^  ?     Is  a^  :  6"^  : :  x?^ 

\  y'-^l  Is  a"*  :  6"*  :  :  a:"'  :  7/"*  whether  m  is  integral  or 
fractional,  positive  or  negative?  AVhy  is  it  that  the 
ratios   remain   equal  in   each   case?      How   are  they 


changed? 


MISCELLANEOUS. 

19.  If  a  :  &  :  :  c  :  (Z,  show  that  ma  -{-  nh  \  pa  -]-  qb  :  :  mc 

-\-  nd  '.  pc  -\-  qd. 

SuG. — The  ratios   to  be  compared  when  reduced  to  a  C.  D.  are, 

acmp  -\-  bcnp  -f-  ndmq  -\-  hdnq  acmp  -f-  bcmq  -\-  adnp  -\-  hdnq 

(Up  -\-bqi{cp  -^dq^  {ap -}- bq}{cp -\- dq) 

Now  from  the  given  proportion  we  learn  that  ad  =  be.  Therefore,  ex- 
changing them  in  the  two  middle  terms  of  the  first  ratio,  the  ratios 
become  identical. 

This  may  also  be  shown  as  follows  :  Multiplying  antecedents  by  m 
and  consequents  by  n,  ma  :  7ib  : :  mc  :  nd.  By  composition  ma  -j-  nb  : 
ma  ::  mc  -{-  nd  :  mc,  or  multiplying  both  ratios  by  m,  ma  -f-  >'^  :  a  :  : 
mc  -{-  nd  :  c.  By  changing  the  places  of  the  means  ma  -f-  nb  :  mc  -\- 
nd  : :  a  :c.  In  like  manner  it  may  be  shown  that  pa-\-qb  '■  pc  -\-  qd  :  : 
a  :c.  .  • .  ma  -{-  nb  :  mc-\-nd  :  :  pa-\-  qb  :pc-\-  qd,  or  ma  -{-  nb  :  pa  -\- 
qb  :  :  mc  -\-  nd  \  \pc  -{-  qd.  The  student  should  give  the  reason  why 
each  step  does  not  vitiate  the  proportion,  according  to  {09). 

20.  If    {a-\-b-^c  +  d){a  —  b  —  c+d)  =  (a  —  b-\-c  —  d) 
(a  -f  6  —  c  —  d)  prove  that  a  :  b  :  :  c  :  d. 

^  ^        ,,       .  ^.  ,        a  +  ?)-j-c-f-d      a  —  b-j-c—d 

buG.  — From  the  given  equation  we  have  — -——^ ■ — -,  = ; — ■ ; — •,. 

a-^b  —  c  —  d      a  —  b  —  c-^d 

Clearing  of  fractions  and  reducing  2ad —  2bc  =  — 2ad  -f-  26c,  or  ad  = 

be.  Whence  -  =  -. 
b  d 
This  may  also  be  proved  by  writing  according  to  Prop.  2,  a  -f-  ^  + 
c-{-d  :  a  —  b  -f-c  —  d  ::  a  +  6  —  c  —  d  :  a  —  6—  c-fd.  Comparing 
the  sum  of  each  antecedent  and  its  consequent  with  their  difference, 
2a  4-  2c  :  26  +  2d  :  :  2a  —  2c  :  26  —  2d  Whence  a  -}-  c  :  a  —  c  : : 
b-\-d  :  6  —  d.     Repeating  the  same  processes  we  have  a  :  6  :  :  c  :  d 

21.  If  — : =  b,    show   that   a  —  x  :  2a   :  :   2b  :  a  -\-  x. 

Produce  other  forms  of  proportion  from  the  given  re- 
lation.    How  many  can  be  produced  ? 


PROPORTION.  293 

22.  If  r=^^s  show  that  r  :  s  :  :  1  :  ^2. 

23.  If  (a  4-  j:)^  :  (a  —  x)^  :  :  x  +  y  :  x  —  y,  show  that  a  : 
X  : :  ^2a  —  y  :  ^^. 

Solution.     a2  -f-  2ax  +  x^  :  a^  —  2ax -{- x^  :  :  x  +  y  :  x  —  y, 


a2 

4-  2ax  +  x2  : 

a2  — 

2ax  4-  x2  : 

2a2  +  2x2 

:  4ax 

:  :  2x  :  2y, 

a^  +  x2  : 

x2  :  : 

2a  :  y. 

a2  :  x2  : : 

2a  — 

y  '■  2/. 

.-.  a  :  X  : 

:  ^/2^ 

—y  :  v/y. 

Let  the  student  give  the  reasons. 

24.  If  a  :  6  : :  c  :  d  ::  e  :f ::  g  :  h  ::  i :  k,  etc.,  show  that 

{a-\-c-\-e  +  g  +  i+,  etc.)  :  (6  +  cZ  -h/4-  h  +  k-^,  etc.) 
:  :  a  :  i*),  or  c  :  c?,  or  e  :  /*,  etc.  That  is,  in  a  series  of  equal 
ratios,  the  sum  of  all  the  antecedents  is  to  the  sum  of  all 
the  consequents,  as  any  antecedent  is  to  its  consequent. 


Solution. 

a       a         ,        - 
-  =  -  OT  ah  =  6a, 
b       b 

a       c          ,       , 

—  =:  -  or  aa  =  he, 
0       a 

-  =  -  OT  af  =  h%, 

a       q         ,        ^  - 
_  =  ^  or  a/.  =  6gr. 

—  =  -  or  a/c  =  hi, 
b        k 

etc. 


Adding,  a(h  -\- d 4-.f  +  h-\^k-\-,  etc. )  =  h{a -\-  c  ^  e -\- g  -{-i  ■\;  etc.) : 
whence  {a-\-c-\-e-{-g-\-i-\-,  etc.)  :  (6  +  d  -j-/  -f-  ^  +  ^  +»  etc. ) 
:  :  a  :  6  or  (since  a  -.b  =  c  -.d,  etc. ),  c  :  d  :  :  e  :/,  etc. 

25.  Four  given  numbers  are  represented  by  a,  b,c,d;  what 

quantity  added  to  each  will  make  them  proportionals  ? 

.  be  —  ad 

Ans,, — •. 

a  —  6  —  c  +  a 

26.  If  four  numbers  are  proportionals,  show  that  there  is 
no  number  which,  being  added  to  each,  wiU  leave  the 
resulting  four  numbers  proportionals. 


294      RATIO,  PROPORTION,  AND  PROGRESSION. 

SECTION  III 
Progressions. 

70,  A  Progression  is  a  series  of  terms  which  in- 
crease or  decrease  by  a  common  difference,  or  by  a  com- 
mon multiplier.*  The  former  is  called  an  Arithmeiical,  and 
the  latter  a  Geometrical  ProgreHsion.  A  Progression  is  in- 
creasing or  Decreasing  according  as  the  terms  increase  or 
decrease  in  passing  to  the  right.  The  terms  Ascending  and 
Descending  are  used  in  the  same  sense  as  increasing  and 
decreasing,  respectively.  In  an  increasing  Arithmetical 
Progression  the  common  difference  is  added  to  any  one 
term  to  produce  the  next  term  to  the  right ;  and  in  a  de- 
creasing progression  it  is  subtracted.  In  an  increasing 
Geometrical  Progression  the  constant  multiplier  by  which 
each  succeeding  term  to  the  right  is  produced  from  the 
preceding  is  more  than  unity;  and  in  a  decreasing  progres- 
sion it  is  less  than  unity.  This  constant  multiplier  in  a 
Geometrical  Progression  is  called  the  Ratio  of  the  series.f 

7-1.  The  character,  •  • ,  is  used  to  separate  the  terms  of  an 
Arithmetical  Progression,  and  the  colon,  : ,  for  a  like  pur- 
pose in  a  Geometrical  Progression. 


*  This  is  the  common  use  of  the  term.  It  is  also  used  to  include  what  is  called  a 
Harmonical  Progression.  But  our  limits  do  not  allow,  neither  does  the  importance 
of  the  subject  demand  the  treatment  of  the  latter  topic  in  this  volume. 

t  This  is  an  unfortunate  use  of  the  term  Ratio  :  it  were  better  to  use  the  term 
Rate,  as  some  French  writers  do.  To  harmonize  the  use  of  the  term  in  proportion, 
with  this  vise,  may  have  led  some  writers  to  define  ratio,  as  used  in  proportion,  as 
the  quotient  of  the  cousi'quent  divided  by  the  antecedent.  But  the  definition  has 
neither  logic  nor  the  common  usage  of  autliors,  English  or  c'oiitinental,  to  support 
it.  How  the  notion  that  this  definition  of  ratio  is  the  common  French  definition, 
has  gained  currency,  it  is  not  easy  to  say.  So  far  as  the  author's  observation  has 
oxteni'^d.  it  certxinlv  i^  not  the  fact. 


ARITHMETICAL    PROGRESSION.  295 


LLLUSTEATIONS. 

1  ..  3  ..  5  ..  7,  etc.,  etc.,  is  an  Increasing  Arithmetical  Progression 
with  a  common  difference  2,  or  -j-  2. 
15  ••  10  ••  5  ••  0  ••  —  5,  etc.,  etc.,  is  a  Decreasing  Arithmetical  Pro- 
gression with  a  conamon  difference  —  5. 
a"a  +  d"a  +  2d"a-\-  3d,  etc. ,  etc. ,  is  the  general  form  of  an 
Arithmetical  Progression,  d  being  the  com- 
mon difference. 
2:4:8  :  16,  etc.,   etc.,  is  an  increasing  Geometrical  Progression 
with  ratio  2. 
12  : 4  :  ^  :  I  :  ^V  e^-.  etc.,  is  a  Decreasing  Geometrical  Progression 

with  ratio  ^. 
a  :  ar  :  ar'^  :  ar^  :  ar*,  etc.,  etc.,  is  the  general  form  of  a  Geometrical 
Progression,  r  being  the  ratio,  and  greater 
or  less  than  unity,  according  as  the  series  is 
increasing  or  decreasing. 


72,  There  are  Five  Things  to  be  considered  in  any  pro- 
gi'ession  ;  viz.,  the  first  term,  the  last  term,  the  common 
difference  or  the  ratio,  the  number  of  terms,  and  the  sum 
of  the  series,  either  three  of  which  being  given  the  other 
two  can  be  found,  as  will  appear  from  the  subsequent  dis- 
cussion. 


ARITHMETICAL  PROGRESSION. 

75.  JPro2>»  1»  The  formula  for  Jinding  the  nth,  or  last 
term  of  an  Arithmetical  Progression;  or,  more  properly,  the 
formula  expressing  the  relation  between  the  first  term,  the  nth 
term,  the  common  difference,  and  the  number  of  terms  of  such 
a  series  ts 

/  =  a  -f  (71  —  l)d, 
in  which  a  is  the  first  term,  d  the  common  difference,  n  the 
number  of  terms,  and  I  the  nth  or  last  term,  d  being  posi- 
tive or  negative  according  as  the  series  is  increasing  or  de- 
creasing. 


RATIO,   PROPORTION,   AND    PROGEESSIOIT. 

Dem. — According  to  the  notation,  the  series  is 

a  •  •  a  -f-  d .  •  a  -f-  2d  •  •  a  -|-  3d . .  a  -f-  4d  •  •  a  -f-  5d,   etc.,   etc. 
Hence  we  observe  that  as  each  succeeding  term  is  produced  by  adding 
the  common  difference  to  the  preceding,  when  we  have  reached  the 
nth  term,  we  shall  have  added  the  common  diflference  to  the  first  term 
n  —  1  times  ;  that  is,  the  nth  term,  or  /  =  a  -f-  (n —  l)d.     q.  e.  d. 

ScH. — As  this  formula  is  a  simple  equation  in  terms  of  a,   I,   n,   and 
d,  any  one  of  them  may  be  found  in  terms  of  the  other  three. 


74.  JProp,  2.  The  formula  for  the  sum  of  an  Arith- 
m^etical  Progression,  or  expressing  the  relation  between  the  sum 
of  the  series,  the  first  term,  last  term,  and  number  of  terms  is 

■a  +  /- 


n^y 


s  representing  the  sum  of  the  series,  a  the  first  term,  I  the 
last  term,  and  n  the  number  of  terms. 

Dem. — If  I  is  the  last  term  of  the  progression,  the  term  before  it  is 
I  —  d,  and  the  one  before  that  I  —  2d,  etc.    Hence,  as  a  •  •  a  +  d  •  •  a  -f- 

2d  ••  a-\-^d I,   represents  the  series,  1--1  —  d--l  —  2d"i!  — 

3d a,  represents  the  same  series  reversed.     Now,  the  sum  of 

the  first  series  is 

s  =  a  -f  (a  -l-d)  +(a+  2d)-\ (Z  _  2d)  +  (?  —  d)  +  Z; 

and  reversed  s  =  I  -f  (I  —  d)  -j-  (^  —  2d)  H (a  +  2d)  -{-  (a  -)-  d)  -f-  a. 

Adding,  2s  =  (a  -|-  /)  -|-  (a  -\-  1} -]-  {a -{- 1)-\ [a-\- 1)  -\-ya  -\-l]-{-{a-\-l}. 

If  the  number  of  terms  in  the  series  is  n,  there  will  be  n  terms  in  this 

sum,  each  of  which  is  {a-\-l);  hence  2s  =  (a  -|-  hn,  or  s  =        7"     p- 

Q.  E.  D. 

ScH. — This  formula  being  a  simple  equation  in  terms  of  s,  a,  I,  and 
n,  any  one  of  the  four  can  be  found  in  terms  of  the  other  three. 

75,  Cor.  1. — Formulas 

(1)  1  =  a  +  (n  —  l)d,         and 

(2)  s=    — TT"  r^'  being   two   equations 

between  the  Jim  quantities,  a,  1,  n,  d,  and  s,  any  two  of  these 
five  can  he  found  in  terins  of  the  other  three. 


ARITHMETICAL  PROGRESSION.  297 

[Note. — It  is  not  considered  worth  while  to  make  separate  cases 
out  of  the  different  problems  which  arise  in  the  progressions,  or  to 
cumber  the  memory  with  the  multipHcity  of  formulas  which  can  ba 
deduced  from  the  two  fundamental  ones,  but  rather  that  these  should 
be  fixed  in  memory,  and  their  use  clearly  understood.  ] 

EXAMPLES. 

1.  The  first  term  of  an  A.  P.  is  2,  and  the  common  differ- 
ence 3,  what  is  the  11th  term  ?  "What  the  sum  of  the 
series  ? 

Solution. — In  the  first  case  there  are  under  consideration  the  first 
term,  a  =  2,  the  common  difference,  d  =  3,  the  number  of  terms, 
n  =  11,  and  the  last  term,  which  is  the  thing  required.  The  relation 
between  these  is  given  in  Z  =  a  -f-  (n  —  l)(i  ;  in  which  by  substituting 
the  given  values  there  results  /  =  2  -f  (11  — 1)3  =  32.   In  the  second  case 

the  formula  s  =  \      J~      n  gives  the  relation,  in  which  by  substituting 
the  given  values  there  results  s  =     — — ^    11  =  187. 

2.  The  first  term  of  an  A.  P.  is  8,  the  last  term  203,  and 
the  common  difference  5,  what  is  the  number  of  terms  ? 
What  the  sum  of  the  series  ? 

Ans.,  n  =  40,  s  =  4220. 

3.  The  first  term  of  an  A.  P.  8,  the  last  term  203,  and  the 
number  of  terms  40,  what  are  the  common  difference 
and  the  sum  ? 

4.  The  last  term  is  1,  the  sum  1717,  and  the  number  of 
terms  34,  what  are  the  first  term  and  the  common 
difference  ? 

SuG.— The  equations  are  1  =  a  +  (34  —  l)d,  and  1717  =  (^-^~-^3iy 
from  which  to  find  a  and  d.     a  =  100,  and  d  =  —  3. 

5.  What  is  the  sum  of  the  numbers  1,  2,  3,  4,  etc.,  to  1000? 
Suggestion. — The  common  difference  is  1,  and  the  last  term  1000. 

6.  The  first  term  of  an  arithmetical  progression  is  1,  and 
the  number  of  terms  23,  what  must  be  the  common  dif- 
ference that  the  sum  of  all  the  terms  may  be  100? 

What  the  last  term? 


298  EATIO,    PROPORTION,  AND    PROGRESSION. 

7.  If  the  first  term  of  an  arithmetical  progression  is  100, 
and  the  number  of  terms  21,  what  must  the  common 
difference  be  that  the  sum  of  the  series  may  be  1260  ? 
What  the  last  term  ?  Ans.,  —  4. 

8.  Two  persons,  A  and  B,  start  from  the  same  place 
together,  and  travel  in  the  same  direction.  A  goes 
40  miles  per  day;  B  goes  20  miles  the  first  day,  and 
increases  his  rate  of  travel  |  of  a  mile  per  day.  How 
far  will  they  be  apart  at  the  end  of  40  days,  and  which 
will  be  in  advance  ?  Ans.  A,  215  miles. 

9.  The  first  term  of  an  arithmetical  progression  is  —  7, 
the  common  difference  —  7,  and  the  number  of  terms 
101,  what  is  the  sum  of  the  series?       Ans.,  —  36057. 

10.  Insert  8  arithmetical  means  between  3  and  21. 

Series.,  5- •  7- -O- -11- -13. -15. -17- -19. 

11.  Insert  3  arithmetical  means  between  i  and  -i-. 

Series,  -3. . .  _5^ . .  li. 

12.  What  is  the  nth  term  of  the  series  l««3-'5««7",  etc. ? 

Ans.,  2n  —  1. 

13.  What  is  the  sum  of  n  terms  of  the  series  l-«3-'5--7'*, 
etc.  ?  Ans.^  n^. 

7G,  Cor.  2. — The  formula  for  inserting  a  given  number  of 

\ a 

arithmetical  meaiis  between  two  given  extremes  is  d= . 

^  m  -h  1' 

in  which  m  represents  the  number  of  means.  From  this  d, 
the  common  difference,  being  found,  the  terms  can  readily  be 
written. 

Dem. — If  a  is  the  first  term  and  I  the  last,  and  there  are  m  terms 
between,  or  m  means,  there  are  in  all  m  -f-  2  terms.  Hence  substituting 
in  the  formula  I  =  a-\-[n  —  l)d  for  n,  m-{-2,  we  have  Z  ■=  a  -f-  ;m  -)-  l)d. 

From  this  d  =  — — -.     q.  e.  d, 
m-{-l 


GEOMETRICAL  PROGRESSION.  299 

14.  If  a  body  falling  to  the  earth  descends  a  feet  the  first 
second,  3a  the  second,  5a  the  third,  and  so  on,  how 
far  will  it  fall  during  the  ^th  second  ? 

Ans.,  {2t—l)a. 

15.  If  a  body  falling  to  the  earth  descends  a  feet  the  first 
second,  3a  the  second,  5a  the  third,  and,  so  on,  how 
far  will  it  fall  in  t  seconds  ?  Ans.,  at\ 

16.  A  debt  can  be  discharged  in  a  year  by  paying  $1  the 
first  week,  $3  the  second,  $5  the  third,  and  so  on  ;  re- 
quired the  last  payment  and  the  amount  of  the  debt. 

Ans.^  Last  payment,  $103  ;  amount,  $2704 

17.  A  person  saves  $270  the  first  year,  $210  the  second, 
and  so  on.  In  how  many  years  will  a  person  who 
saves  every  year  $180  have  saved  as  much  as  he  ? 

Ans.,  4. 

18.  A  board,  2|  inches  wide  at  the  narrow  end,  and  10  feet 
long,  increases  in  width  1\  inches  for  every  foot  in 
length  ;  what  is  the  width  of  the  wide  end  ? 

Ans.,  17|  in. 

19.  If  100  oranges  are  placed  in  a  line,  exactly  2  yards  from 
each  other,  and  the  first  2  yards  from  a  basket ;  what 
distance  must  a  boy  travel,  starting  from  the  basket,  to 
gather  them  up  singly,  and  return  with  each  to  the 
basket?  Ans.,  11  mi.  3  fur.  32  rd.  4  yd. 

[Note. — For  other  examples  involving  the  principles  of  Arithmetical 
Progi-ession,  see  Problems  after  Quadratics,  and  also  the  subject  of 
Interest.] 


GEOMETRICAL   PROGRESSION. 
77.  I^rop,  1.   The  formula  for  finding  the  nth,  or  last 
term,  of  a  Geometrical  Progression;   or,  more  properly,  the 
formula  expressing  the  relation  between  the  first  term,  the  nth 


300  RATIO,  PROPORTION,  AND   PROGRESSION. 

term,  the  ratio,  and  the  numher  of  terms  of  such  a  series  is 
l=ar"~*,  in  which  1  is  the  last,  or  nth  term,  a  the  first  term, 
r  the  ratio,  and  n  the  numler  of  terms. 

Dem. — Letting  a  represent  the  first  term  and  r  the  ratio,  the  series 
is  a  :  ar  :  ar^  :  ar'^  :  ar^  :  etc.  Whence  it  appears  that  any  term  con- 
sists of  the  first  term  multiplied  into  the  ratio  raised  to  a  power  whose 
exponent  is  one  less  than  the  number  of  the  term.  Therefore  the  nth 
term,  or  I  =  ar''-'^.     q.  e.  d. 

7S,  I^vop*  2,  The  formula  for  the  sum  of  a  Geometrical 
Progression,  or  expressing  the  relation  between  the  sum  of  the 
series,  the  first  term,  the  ratio,  and  the  number  of  terms  is 

ar" —  a 

in  which  s  represents  the  sum,  a  the  first  term,  r  the  ratio, 
and  n  the  number  of  terms. 

Dem. — The  sum  of  the  series  being  found  by  adding  all  its  terms, 
we  have, 

s=a  -|-ar-j-ar2-j-ar3  -j ar"~^-\-ar^  -^-f-ar"— i,    and  multiplying 

by  r,  rs  =      ar-\-ar^-\-ar^  -| ar"-^-\-ar''-^-\-ar''-^-\-ar".  Subtracting, 

rs  —  s  =  ar"  —  a,     or 
(r  —  l)s  =  ar"  —  a,  and    s  = —     q.  e.  d. 

SuG.  —The  student  will  notice  that  multiplying  the  first  series  by 
r,  and  placing  the  terms  of  the  product  under  the  like  terms  of  the  se- 
ries, simply  moves  each  term,  when  multipUed,  one  place  to  the  right, 
so  that  however  many  terms  there  may  be  in  the  series,  each  will  have 
a  similar  one  in  the  product  except  the  first  term,  a  ;  and  each  term 
in  the  product  wiU  have  a  similar  one  in  the  series,  except  the  last  one, 


70,  Cor.  1. — Formulas 

(1)  1  =  ar— S  and 

(2)  s  =  — ; -—  being  two  equations 

between  the  five  quantities,  a,  1,  r,  n,  and  s,  are  sufficient  to  de- 
termine any  two  of  them  when  the  others  are  given. 


GEOMETRICAL  PROGRESSION.  SOI 

S0»  Cor.  2. — Since  1  =  ar"~\  Ir  =  ar",  which  substituted 
in  (2)  tfives  s  = ;  which  formula  is  often  convenient. 

EXAMPLES. 

1.  The  first  term  of  a  geometrical  progression  is  2,  the  ra- 
tio 3,  and  the  number  of  terms  6.  What  are  the  last 
term  and  the  sum  of  the  series  ? 


r  —  1 

2.  The  last  term  of  a  geometrical  progression  is  62500, 
the  ratio  5,  and  the  number  of  terms  7.  What  are  the 
first  term  and  the  sum  of  the  series  ? 

SuG. — From  I  =  ar"-^,  tliere  results  by  substitution  62500  =  a-5^ 

or  15625a  =  62500.     .  • .  a  =  4.     From  s  =         ~  "^  we  find  s  = 

r  —  1 

78124. 

3.  By  saving  1  cent  the  first  week,  2  cents  the  second 
week,  4  cents  the  third  week,  and  so  on,  doubhng  the 
amount  every  week,  how  much  is  saved  the  last  week 
of  the  year  ?  A7is.,  $22,517,998,136,852.48. 

SuG.— This  problem  requires  us  to  raise  2  to  the  51st  power.  This 
is  readily  effected  thus  :  the  3rd  power  of  2  is  8.  The  3rd  power  mul- 
tiplied by  the  3rd  power  gives  the  6th  power;  hence  8  X  8  =  64  is  the 
6th  power.  In  like  manner  64  X  64  =  4096  is  the  12th  power,  and 
4096  X  4096  =  16777216  is  the  24th  power;  and,  finally,  16777216  X 
16777216  X  8  is  the  51st  power. 

4.  What  is  the  sum  of  10  terms  of  the  series  8:4:2: 
1  :  i,  etc.  ? 


1)       'ih-') 


a(r--l)        "V2'"           /        ^,    2'"  -  1        1023 
SUG.     s=  —-^  =  __^ =  2.  -^^^  =-^    =. 

2 

^  =  15tJ.      Also,  Z^saf-i  =    -==-  =-^4-.     In  such  cases  the 

ingenious  student  will  avoid  unnecessary  multipUcations  by  suppress- 
ing faetors. 


302  RATIO,   PROPORTION,   AND  PROGRESSION. 

5.  If  4  is  the  first  term,  324  the  last,  and  5  the  number  of 
terms,  what  is  the  ratio  ? 

SuG. — Since  I  =  ar''-^,  we  have  324  =  4r*,  or  r  =  ^^81  =  3. 

6.  Insert  5  geometrical  means  between  3  and  192. 

The  ratio  is  2. 

7.  Find  4  geometrical  means  between  -^  and  ^. 

(2\  ^ 
-  )   ,   whence  the   series  becomes 

11  1  1  11 


mtans  between  a  and  1  is 


9'         4  1'3  2*2  3'1  4''^ 

"^      2^-3^      2^  •  3"^      2^  •  3"^      2^  •  3'^      *^ 
81*  CoR.  3. — The  formula  for   inserting   m   geometrical 

fi 

IS  r  =  "^+  M--. 
\a 

ScH. — This  and  many  other  problems  in  Geometrical  Progression, 
are  more  readily  solved  by  means  of  logarithms.  Many  also  require 
a  knowledge  of  quadratic  equations,  and  even  of  the  higher  equations. 
Some  farther  illustrations  will  be  given  in  their  proper  place,  espe- 
cially in  treating  the  subject  of  Interest. 

82,  CoR.  4. — The  formula  for  the  sum  of  an  Infinite  De- 
creasing Geometrical  Progression  is  s  = ;. 

Dem. — Since  in  a  decreasing  progression  the  ratio  is  less  than  unity, 
the  last  term,  ar'^"^,  is  also  less  than  the  first  term,  and  numerator  and 

denominator  of  the  value  of  s,  •- ,   become    negative.     Hence  it 

is  well  enough  to  write  the   formula  for   the  sum  of  such  a  series 

s  = ,  that  is,  change  the  signs  of  both  terms  of  the  fraction. 

1  —  r 

Now,  if  the  terms  of  a  series  are  constantly  decreasing,  and  the  num- 
ber of  terms  is  infinite,  we  can  fix  no  value,  however  small,  which  will 
not  be  greater  than  the  last,  or  than  some  term  which  may  be  reached 
and  passed.     Hence  we  are  compelled  to  call  the  last  term  of  such  a 

series  0,  which  makes  the  formula  s  =  — — .     q.  e.  d. 


GEOMETRICAL  PROGRESSION.  303 

ScH.— Decimal  Repetends  afford  illustrations  of  such  series.  Thus 
.333  +,  is  ,%  +  r^u  +  TJlo  +,  etc.,  to  infinity.  Again,  .5434343  -f-  is, 
fj  -f  the  series  t^^o  +  To^gotr  +Ttrtr^^ot7  +>  etc.,  to  infinity. 

8.  Find  the  sum  of  the  series  1  :  ^  :  i  :  etc.,  to  infinity. 

Sum,  2. 

9.  Required  the  sum  of  the  series  1,  i,  -^j to  infinity. 

Sum,  li. 

10.  Find  the  value  of  .1212 to  infinity.        Sum,  ^V 

11.  Find  the  value  of  .2333,  etc.,  to  infinity.        Sum,  ^V 

12.  Find  the  value  of  .3411111,  etc.,  to  infinity. 

Sum,  ^U- 

13.  Find  the  value  of  .323232,  etc.,  to  infinity. 

Sum,  ^i. 

14    Find  the  value  of  .20414141,  etc.,  to  infinity. 

Sum,  ^U- 

15.  Suppose  a  body  to  move  eternally  in  this  manner;  viz., 
20  miles  the  first  minute,  19  miles  the  second  minute, 
18^  the  third,  and  so  on  in  geometrical  progression. 
What  is  the  utmost  distance  it  can  reach  ? 

Ans.,  400  miles. 

16.  What  is  the  distance  passed  through  by  a  ball,  before 
it  comes  to  rest,  which  falls  from  the  height  of  50  feet, 
and  at  every  fall  rebounds  half  the  distance  ? 

Ans.,  150. 

17.  In  the  preceding  problem,  suppose  the  body  falls 
16yL.  feet  the  first  second,  3  times  as  far  the  next 
second,  and  5  times  as  far  the  third  second,  and  so 
on,  how  long  will  it  be  before  it  comes  to  rest? 

Ans.,  -h%-  v/579(4  +  3^2)  =  10.27657+  seconds. 


304 


RATIO,   PROPORTION,  AND  PROGRESSION. 


Synopsis. 


o 
o 

I 

o 


t 


{Ratio. — Terms  of. — Antecedent. — Consequent.—' 
Couplet. 
Direct. — Inverse. — Gr.  inequality,  less. 
Comp.  ratio. — Duplicate,  sub-duplicate,  etc.,  etc. 

Sign  of. 

Cor. — Changes  in  terms  of. 

r  Proportion. — Extremes. — Means. 

J  Mean  proportional. — Third  proportional. 

1  Inversion.  — Alternation.  — Composition,  — Division, 

[inverse  or  reciprocal  proportion. — Continued. 

{Pro]).  l.—Cor.  l.—Cor.  2. 

J  Prop.  2. 

1  Prop.  3. — Principles  on  which  transformations  are 

[  made. 

Equi-multiples. — Why  proportion  not  destroyed, 
dig.  in  order  of  terms, —      "  "  " 

'Composition,  or  division. —   "  *'  " 

Involution,  or  evolution. —  "  "  " 

Progression.  — Arithmetical.  — Geometrical. 
Increasing,  or  ascending, — Decreasing,  or  descend- 
ing. 
Common  difference,  positive,  negative. 
,  Ratio,  greater  than  1,  less  than  1. 

Sign  of  Ar.  Prog. — Of  Geometrical. 

Pive  things. — Given,  required. 

Produce    them. 


2    Fund,    formulas 
means. 


-To    insert 


2  Fund,  formulas. — Pro-  )  To  insert  means. 

duce  them.  s  Sum  of  infinite  series. 


Test  Questions. — Give  the  various  changes  which  can  be  made 
upon  the  terms  of  a  ratio  and  tell  how  the  ratio  is  aflfected,  and  Why  ? 
State  the  various  transformations  which  can  be  made  upon  a  propor- 
tion without  destroying  it,  and  Give  the  reason  in  each  case.  Produce 
the  two  fundamental  formulas  of  Arithmetical  Progression.  Also  of 
Geometrical. 


APPLICATIONS.  305 


APPLICATIONS. 

[Note. — ^Teacher  and  pupil  should  bear  in  mind  that  the  object  of 
this  section  is  to  teach  the  properties  of  Ratio  and  Proportion  ;  hence 
all  the  operations  should  be  performed  upon  the  proportion.  The  pro- 
portion should  be  kept  in  the  form  of  a  proportion,  and  not  reduced 
to  an  equation.] 

1.  Divide  60  into  two  parts  which  are  to  each  other  as  2  :  3. 

SuG. — Letting  x  and  60  —  a;  be  the  parts,  ic  :  60  —  x  :  :  2  :  3.  Hence 
x  :  60  :  :  2  :  5,  or  X  :  24  :  :  2  :  2  ;  and  x  =  24.  The  pupil  should  give 
the  reason  for  each  transformation.  What  is  the  first  transformation  ? 
Composition.  Why  does  it  not  destroy  the  proportion?  What  the 
second  transformation  ?     Why  does  it  not  destroy  the  proportion  ? 

2.  A  boy  being  asked  his  age  said  :  John,  who  is  18,  is 
older  than  I  ;  but,  if  you  add  to  my  age  ^  of  it,  and 
from  this  sum  subtract  ^  of  my  age,  the  result  will  be 
to  John's  age  as  10  :  9.     How  old  was  the  boy  ?    Verify. 

OPEEATioN.     X  -\-  ix  —  4  X  :  18  : :  10  :  9, 
f  X  :  18  :  :  10  :  9, 
X  :  18  :  :  8  :  9, 
X  :  18  :  :  16  :  18, 
.-.  x  =  16. 
Let  the  pupil  tell,  in  each  instance,  just  what  the  transformation  is, 
and  why,  according  to  (60),  the  proportion  is  not  destroyed. 

Vebification.  16  -|-  -•/  —  ^£-  =  20,  which  is  to  18  as  10  is  to-9,  the 
ratio  in  case  being  ^^. 

3.  Two  brothers  being  asked  their  ages,  the  younger  re- 
plied, my  age  is  to  my  brother's  as  2  to  3  ;  and  if  you 
add  18  to  mine  and  2  to  his,  the  sums  will  be  as  3 
to  2.     What  were  their  ages  ? 

SuG. — To  solve  with  one  unknoASTi  quantity  we  may  represent  the 
younger  brother's  age  by  2x  and  the  older's  by  3x. 
Then       2x  +  18  :  3x  -f  2  :  :  3  :  2  ; 
Whence  2x  -f  18  :  27x  +  18  :  :  3  :  18, 

2x  +  18  :  25x  :  :  3  :  15  :  :  1  :  5, 
2x  -f  18  :  X  :  :  5  : 1, 
2x  +  18  :  2x  :  :  5  :  2, 
18  :  2x  :  :  3  :  2, 
9  :  X  :  :  9  :  6, 
.  • .  X  =  6,  2x  =  12  and  3x  =  18. 


306  PROPORTION. 

[Note. — True,  it  gives  a  somewhat  shorter  solution  of  this  example 
to  put  the  first  proportion  immediately  into  the  equation  4x  -j-  36  = 
9x  +  6,  whence  5x  =  30  and  tc  =  6.  But  the  object  is  to  become 
famiUar  with  the  properties  of  a  proportion.  ] 

4.  A  man's  age  when  lie  was  married  was  to  his  wife's  as 
3  to  2  ;  but  after  4  years,  his  age  was  to  hers  as  7  to  5. 
What  were  their  ages  when  they  were  married  ? 

SuG. — This  may  be  solved  with  one  unknown  qiiantity,  like  the  pre- 
ceding, and  that  is  the  more  elegant  way.  We  may  also  use  two. 
Thus,  X  :y  :  :3  :2,  and  x  +  4  :  j/  +  4  :  :  7  :  5.  From  the  former 
X  :  f  i/ :  :  3  : 3  .  • .  x  =  fy.  Substituting  in  the  latter  f  y  -|-  4  :  r/  -}-  4  :  :  7  : 5. 
Whence  32/  +  8  •.2y-\-8  :  :  7  :  5,  y:2y-i-S  :  :  2  :  5.  2^/  :  2y  +  8  :  :  4  :  5, 
2?/  :8  :  :4  :  1,  andy  :  1  :  :16  :1.     .-.  y=16,  andx  =  fy=24. 

5.  A  man  is  now  25  years  old  and  his  brother  is  15.  How 
many  years  before  their  ages  will  be  as  5  to  4  ?     Verify. 

6.  A  man  has  two  flocks  of  sheep,  each  containing  the 
same  number;  from  one  he  sold  80  and  from  the  other 
20,  when  the  numbers  in  the  flocks  were  as  2  to  3 
How  many  were  there  in  each  flock  in  the  first  place  ? 
Verify. 

7.  It  is  known  to  every  one  that  a  small  body  near  the 
eye  hides  a  large  one  farther  off;  and  it  is  a  principle 
in  optics  so  nearly  axiomatic  that  we  will  take  it  for 
granted,  that,  in  order  to  have  the  smaller  body  just 
cover  the  larger  their  distances  from  the  eye  must  be 
in  proportion  to  their  breadths,  or  lengths.  From  this 
some  very  pleasing  calculations  can  be  made.  The 
pupil  may  make  the  following  : 

1st.  The  breadth  of  a  man's  thumb  is  about  1  inch, 
and  he  can  readily  hold  it  at  2  feet  from  his  eye  ;  how 
far  off  is  the  man  who  is  5  feet  8  inches  high,  when  the 
breadth  of  the  man's  thumb  at  2  feet  from  his  eye  just 
covers  the  man?  Arts.,  136  feet 


APPLICATIONS. 


307 


2nd.  Wishing  to  know  approximately  the  height  of  the 
top  of  a  steeple  from  the  ground,  I  found  that  my 
hand,  which  is  4  inches  wide  near  the  thumb,  when 
held  2  feet  from  my  eye,  just  covered  the  height  of  the 
steeple  at  a  distance  of  240  paces  of  3  feet  each.  What 
was  the  height  of  the  top  of  the  steeple  from  the  ground? 

Ans.,  120  feet. 

8.  What  number  is  that  to  which  if  1,  5,  and  13  be  seve- 
rally added,  the  second  sum  will  be  a  mean  propor- 
tional between  the  other  two  ?     Verify. 

9.  What  number  is  that  whose  ^  increased  by  2  is  to  its 


10. 


i  diminished  by 


as  6  is  to  2^. 


Ans.,  30. 


The  number  of  acres  a  farmer  planted  with  corn  is  to 
the  number  he  planted  with  potatoes,  as  f  to  1 ;  but  if 
he  had  planted  6  acres  less  of  corn,  and  ^  as  many  po- 
tatoes +  15^  acres,  the  ratio  would  have  been  as  f  to 
I".     How  many  acres  of  each  did  he  plant  ? 


[Note. — The  five  following  examples  are  designed  to  be  solved  by 
using  two  or  more  unknown  quantities.  ] 

11.    Find  two  numbers,  the  greater  of  which  shall  be  to  the 
less,  as  their  sum  to  42 ;  and  as  their  difference  is  to  6. 


SuG.                       Let  X 

=  one  and  y  the  oth 

Then,                (1) 

X 

:  y  :  :  X  -j-  y  :  4:2, 

02) 

X 

:  y  ::  X  —  y  :    6, 

By  equality  of  ratios 

X 

+  y  :A2::x~  y 

and 

X 

-\-  y  -x  —  y  ::  7 

2x  :2y  ::  8  :  6, 
X  :2y  ::4:  :6, 
x  :  f  y  :  :  4  :  4, 

.-.     X  =  ^  y. 

Substituting  in  (2) 

1?/  ■  y  •■  3.y  —  y 

Or, 

1     :  1   :  :  iV  :     6, 

4     :  1   ::    y  :     6 

1     :  1   :  :    y  :  24: 

■■•     2/  =  24, 

and 

x=^y  =  32. 

:  6, 


308  PKOPORTION. 

12.  Two  numbers  have  such  a  relation  to  each  other,  that 

if  4  be  added  to  each,  they  will  be  in  proportion  as  3 

to  4;  and  if  4  be  subtracted  from  each,  they  will  be  to 

each  other  as  1  to  4.     What  are  the  numbers  ? 

Ans.j  5  and  8. 

SuG.     X  -{-  4:  :  y  -{- 4:  ::3:4  and  x  —  4:1/  —  4::  1:4.     Whence 

-f  1  :  3  :  :  1/  :  4,  or  —~  :  1  :  :  y:l.     .-.  y=  — ^.  Substitut- 
o  o 

ing,x  — 4  : — J^-— 4  :  :  1  :4,  a  — 4  :  2a;  — 4  :  :  ]  :6,  x  — 4  :  x  :  :  1  :5, 

4  :  X  :  :  4  :  5.     .  •.  x  =  5.     Substituting,  6  :  3  :  :  t/  :  4,  or  3  :  3  :  :  y  :  8. 

.-.2/ =  8. 

13.  Find  two  numbers  in  the  ratio  of  2|-  to  2,  such  that, 
when  each  is  diminished  by  5,  they  shall  be  in  the  ratio 
of  1^  to  1.  Numbers,25  and  20. 

14.  There  are  two  numbers,  which  are  to  each  other,  as  16 
to  9,  and  24  is  a  mean  proportional  between  them. 
What  are  the  numbers  ?  Ans.y  32  and  18. 


[Note. — The  following  examples  may  be  solved  by  converting  the 
proportion  into  an  equation,  at  whatever  stage  of  the  solution  it  is 
found  expedient.  ] 

15.  Find  two  numbers  in  the  ratio  of  5  to  7,  to  which  two 
other  required  numbers  in  the  ratio  of  3  to  5  being  re- 
spectively added,  the  sums  shall  be  in  the  ratio  of  9  t,3 
13;  and  the  difference  of  those  sums  =  16. 

Numbers,  30  and  42,  and  6  and  10. 

16.  A  farmer  hires  a  farm  for  $245  per  annum;  the  arable 
land  being  valued  at  $2  an  acre,  and  the  pasture  at 
$1.40;  now  the  number  of  acres  of  arable  is  to  half  the 
excess  of  the  arable  above  the  pasture  as  28  :  9.  How 
many  acres  are  there  of  each  ? 

Ans.,  98  acres  of  arable,  and  35  of  pasture. 

17.  The  quantity  of  water  which  flows  from  an  orifice  is 
proportioned  to  the  area  of  the  orifice,  and  the  velocity 


APPLICATIONS.  309 

of  the  water.  Now  there  are  two  orifices  in  a  reservoir, 
the  areas  being  as  5  to  13,  and  the  velocities  as  8  to  7, 
and  from  one  there  issued  in  a  certain  time  561  cubic 
feet  more  than  from  the  other.  How  much  water  did 
each  orifice  discharge  in  this  time  ? 

Ans.,  440  and  1001  cubic  feet. 

X8.  At  an  election  for  two  members  of  parliament,  three 
men  offer  themselves  as  candidates,  and  all  the  elect- 
ors give  single  votes.  The  number  of  voters  for  the 
two  successful  ones  are  in  the  ratio  of  9  to  8;  and  if 
the  first  had  had  seven  more,  his  majority  over  the 
second  would  have  been  to  the  majority  of  the  second 
over  the  third  as  12  ;  7.  Now  if  the  first  and  third 
had  formed  a  coahtion,  and  had  one  more  voter,  they 
would  each  have  succeeded  by  a  majority  of  7.  How 
many  voted  for  each  ? 

Ans.,  369,  328,  and  300,  respectively. 

19.  A  man,  driving  a  flock  of  geese  and  turkeys  to  market, 
in  order  to  distinguish  his  own  fi'om  any  he  might  meet 
on  the  road,  pulled  5  feathers  out  of  the  tail  of  each 
turkey,  and  2  out  of  the  tail  of  each  goose,  and  upon 
counting  them,  found  that  the  number  of  turkeys' 
feathers  lacked  15  of  being  twice  those  of  the  geese. 
Having  bought  20  geese  and  sold  15  turkeys  by  the  way, 
he  found  that  the  number  of  geese  was  to  the  number 
of  turkeys  as  8  to  3.  What  was  the  number  of  each 
at  first?  Ans.y  45  turkeys,  and  60  geese. 


PROBLEMS  OF  PURSUIT. 

20.  A  fox  starts  up  120  feet  ahead  of  a  hound  at  exactly 
■i-  past  2  o'clock  P.  M.;  the  hound  gives  chase  and 
gains  5  feet  every  2  minutes.  At  what  time  will  he 
overtake  the  fox  ? 


310  PEOPORTION. 

Statement. — Lotting  x  be  the  time  which  will  elapse  before  the 
hound  overtakes  the  fox,  the  problem  becomes;  If  a  hound  gain  5  feet 
in  2  minutes,  how  long  will  it  take  him  to  gain  120  feet?  That  is 
5  :  120  :  :  2  :  a;.  .  •.  x  =  48,  and  the  hound  overtakes  the  fox  at  3 
o'clock  and  18  minutes. 

21.  A  privateer  espies  a  merchantman  10  miles  to  lee- 
ward at  11.45  A.  M.,  and,  there  being  a  good  breeze, 
bears  down  upon  her  at  11  miles  per  hour,  while  the 
merchantman  can  only  make  8  miles  per  hour  in  her 
attempt  to  escape.  After  2  hours  chase  the  topsail  of 
the  privateer  being  carried  away,  she  can  only  make 
17  miles  while  the  merchantman  makes  15.  At  what 
time  will  the  privateer  overtake  the  merchantman  ? 

Ans.,  At  5.30  P.  M. 

22.  A  hare,  50  of  her  leaps  before  a  greyhound,  takes  4 
leaps  to  the  greyhound's  3;  but  2  of  the  greyhound's 
leaps  are  as  much  as  3  of  the  hare's.  How  many  leaps 
must  the  greyhound  take  to  overtake  the  hare? 

SuG.  Let  3x  =  the  number  of  the  hound's  leaps, 

whence  4a;  =     "  "  "      hare's         " 

in  the  same  time.     Then  2:3  :  :  3x  :  4iC  -J-  50.       .•.  x  =  100. ;  and  the 
hound  takes  300  leaps. 

23.  The  hour  and  minute  hands  of  a  clock  are  exactly  to- 
gether at  12  M.     When  are  they  next  together  ? 

SuG. — Measuring  the  distance  around  the 
dial  by  the  hour  spaces,  the  whole  distance 
around  is  12  spaces.  Now,  when  the  hour 
hand  gets  to  1,  the  minute  hand  has  gone 
clear  around,  or  over  12  spaces.  But  as  the 
hour  hand  has  gone  one  space,  the  minute 
hand  has  gained  only  11  spaces.  Now  as 
the  minute  hand  must  gain  an  entire  round, 
or  12  spaces,  to  overtake  the  hour  hand,  we 
have  the  question:  If  the  minute  hand  gains 
11  spaces  in  1  hour,  how  long  will  it  take  to 
gain  12  spaces?  .-.  11  :  I'i  :  :  1  hour  :  x 
hours  ;  and  x  ' 

minutes. 


APPLICATIONS.  311 

ScH. — 1.  It  is  evident  that  the  hands  are  together  every  l-j^,-  hours  ; 
hence  to  find  at  what  time  they  are  together  between  any  two  hours  on 
the  dial,  we  have  only  to  multiply  l-,J-i-  by  the  number  of  the  whole 
hours  past  12  o'clock.  Thus  between  7  and  8  they  pass  each,  other  at 
7-n-,  or  7  o'clock  and  38-i\  minutes.  Between  10  and  11  they  pass  each 
other  at  10  o'clock  o4-i^f  minutes. 

24.  At  what  time  between  6  and  7  o'clock  is  the  minute 
hand  just  i  of  the  circle  in  advance  of  the  hour  hand? 

SuG, — The  question  is  :  If  the  minute  hand  gains  11  spaces  in  one 
hour  how  long  will  it  take  it  to  gain  64  rounds,  or  75  spaces  ?  Or,  if  it 
gains  1  round  in  l/j-  hours,  how  long  will  it  take  it  to  gain  64  rounds? 

Ans.,  At  49-jL  minutes  past  6. 

25.  At  what  time  between  4  and  5  is  the  hour  hand  of  a 
watch  just  20  minutes  in  advance  of  the  minute  hand  ? 
Ans.,  At  no  time  between  these  hours.    The  minute  hand 

is  within  20  minutes  of  the  hour  hand  at  4  o'clock, 
and  at  5y\-  minutes  past  5. 

26.  Before  noon,  a  clock  which  is  too  fast,  and  points  to 
afternoon  time,  is  put  back  5  hours  and  40  minutes  ; 
and  it  is  observed  that  the  time  before  shown  is  to  the 
true  time  as  29  to  105.     Required  the  true  time. 

Si'G. — Letting  x  =  the  time  pointed  to,  x  :  x  +  C)^  :  :  29  :  105.  Observe 
that  to  turn  the  hands  back  6h.  40m.  is  the  same  as  to  turn  them  for- 
ward 6h.  20m.  X  =  2h.  25m.,  and  the  true  time  was  2h.  26m. +  6h.  20m., 
or  15m.  before  nine  o'clock  in  the  morning. 

27.  Two  bodies  move  uniformly  around  the  circumference 
of  the  same  circle,  which  measures  s  feet.  When  they 
start,  one  is  a  feet  before  the  other  ;  but  the  first  moves 
m  and  the  second  31  feet  in  a  second.  When  will  these 
bodies  pass  each  other  the  1st  time,  when  the  2nd, 
when  the  3d,  etc.,  supposing  that  they  do  not  disturb 
each  other's  motion,  and  go  around  the  same  way  ? 

SuG. — 1st.  If  J/>>  m,  the  second  gains  M —  m  feet  a  second,  and 

having  a  feet  to  gain,  overtakes  the  first,  and  does  it  in sec- 

M  —  m 
onds.     The  problem  is  then  like  the  preceding,  as  the  second  gains  a 


312  PROPORTION. 


whole  round  every  — seconds.     Hence  the  second  passing  is  at 

from  the  starting,    the   third  at  — ,    the    fourth    at 


M  —  m  °  M 

a-f3s 


M 


etc. 


2nd.   If  Jf  <<  m,  the  second  is  over- 
taken by  the  first  after  the  first  has  gained 

s  —  a  feet,  or  in —  seconds ;  and  in 

m  —  M 
g 
every —    seconds     thereafter ;    that 

2  o  n 

is,  from  the  time  of  starting,  in  — . 

TO  —  M 
3s  —  a 

etc. 


m  —  M 

28.  When  will  they  pass  if  the  1st  starts  t  seconds  before 
the  second,  and  M  ^  m?    When  if  M  <^m? 

Ans.,  If  M^m,  in  — ,  — ^— ,  —^ — , 

etc.,  seconds.    If  Jf  <"  m,  m ^^j^,  — -, 

3s  —  a  —  TYit     .  T 
-— -,  etc.,  seconds. 

[Note. — Observe  that  the  results  which  are  most  symmetrical  in  the 
two  results  do  not  correspond  to  the  same  meetings.  Thus  the  num- 
ber of  the  time  of  meeting  in  the  first,  is  one  ahead  of  the  coefficient  of 
s.  The  time  of  the  third  meeting  has  2s,  etc.  But  in  the  second  the 
times  of  meeting  and  the  coefficients  of  s  are  the  same. 

29.  When  will  they  pass  if  the  first  starts  t  seconds  later 

than  the  second  and  M'^  m?     When  if  M  <^m? 

^     a~Mt  s-}- a—Mt  2s-j-a—Mt     ^ 
Ans.,  In  -T-p ,  — r— ,  — ,  etc.,  seconds, 

.     s—  a  +  3ft     2.S'  — a  +3fl;     ds  —  a-^3ft 

.  ^^  ^^  -^^:z:  M  '     m  —  31    '  ~^^r=:  jf   *  ^^"^-^ 

seconds. 

30.  When  will  they  meet  if  they  start  at  the  same  time  and 
move  towards  each  other,  or  over  the  distance  a,  first  ? 


APPLICATIONS.  313 

If  they  move  from  each  other,  or  over  the  ares  — a  first  ? 

Ans.,  In  -77— ,  —p ,  -^rp- — ,    etc.,   seconds,  or  in 

s  —  a    2s  —  a    3s  —  a     ,  ^ 

etc.,  seconds. 


M  -\-m'  M-\-  rn   M  +  m 

31.  When  will  they  meet  if  the  first  starts  t  seconds  before 

the  other,  and  they  move  toward  each  other,  or  over 

the  distance  a  first  ?     If   they  move  from  each  other, 

or  over  the  arc  s  —  a  first  ? 

_    s  —  a  —  mt      2s  —  a  —  m^   3s  —  a  —  mt      , 
Ans.,  In  — TT ,     rp— ,  — iT^- ,  etc., 

seconds  after  the  second  starts,  if  they  move  over  the 
arc  s  —  a  first. 

32.  If  they  move  in  opposite  directions,  and  the  first  starts 

t  seconds  later  than  the  second  :  when  they  move  over 

the  arc  a  first  ?    When  they  move  over  the  arc  s  —  a 

first? 

a  —  3It  s  +  a  —  3It  2s  +  a  ~  3It  Ss+a—M/    ^ 
Ans.,  1st.  ^rp- ,  — ^rr— , -— ,  — 5rj-- ,  etc. 

seconds  from  the  time  the  first  starts,  or 
a  -{-  mt     s  -\-  a  -{-  mt  ^s  -}-  a  -}-  mt 


M-\-m  M -\- m  M^m 

from  the  time  the  second  starts. 
2nd.   [Let  the  pupil  determine.] 


etc., 


ScH, — These  problems  are  of  little  value  for  the  purpose  of  illustrat- 
ing the  use  of  the  equation,  but  yet  will  be  found  well  adapted  to  ini- 
tiate the  pupil,  into  the  method  of  reasoning  by  means  of  general  sym- 
bols. The  pupil  should  exercise  his  ingenuity  in  ascertaining  how 
many  different  cases  these  problems  give  rise  to.  Thus,  1st  Class  of 
cases,  when  both  move  in  the  same  direction.  This  will  be  subdivided 
into  (A),  when  they  start  together,  and  {B)  when  they  do  not;  the  lat- 
ter of  which  will  comprise  two  cases,  (a)  when  the  first  starts  first,  ( h) 
when  the  second  starts  first.  Finally,  each  of  these  will  have  iwo  varie- 
ties depending  upon  whether  3f  ^  m,  or  if  <<  m. 


314  BUSINESS  RULES. 

CHAPTER  m. 

BUSIJ^ESS  RULES  [OF  ARITHMETIC^* 


SECTION  L 

Percentage. 

83.  According  to  our  definition,  the  Equation,  of  which 
it  is  the  special  province  of  Algebra  to  treat,  is  the  grand 
instrument  for  investigating  the  relations  of  quantities. 
Now,  in  simple  Percentage,  there  are  four  quantities  to  be 
compared  ;  viz.,  the  Base,  the  Rate  Per  Cent.,  the  Percentage, 
and  the  Amount,  and  the  problem  is.  To  discover  and  ex- 
press in  equations  the  relations  between  these  four  quan- 
tities so  that  if  a  sufficient  number  of  them  are  given  the 
others  may  be  found. 

[Note.  —  For  Definitions  see  Elements  of  Arithmetic,  235 
et  seq.l^ 

84,  I^TOh,  1,  To  express  the  relation  between  base,  rate 
per    cent.,  and  percentage. 

Solution. — Let  b  represent  the  base,  r  the  %,  and  p  the  percentage. 
Now,  if  we  divide  the  base,  h,  by  100,  we  get  1  of  every  100,  or  — . 

But  r%  means  r  of  every  hundred  of  the  base.     Hence  r%  of  6  is  r 

h  rb  rb 

times  J- ,    or     _.     ■■ .  p  =  ^. 

86.  I*rob.    2.  To  express  the  relation  between  rate  per 
cent.,  amount  or  difference,  and  base. 

Solution. — Let  s  represent  the  sum  or  difference  of  the  base  and 
percentage.     Thens  =  &4.  ,^  = ]m~~'   *^®  +  ^^^°  *°  ^^  ^^^^ 


PERCENTAGE.  315 

when  the  base  is  increased  by  the  percentage,  and  the  —  sign  when  the 
base  is  diminished  by  the  percentage. 
ScH. — The  two  formulce 

(1)    p  =  j^    and 

._100i^) 
"^   '  lUU 

expressing  the  relation  between  the  four  quantities  h,  r,  p,  and  5,  two 
of  which  must  always  be  given  to  find  the  others,  are  in  themselves 
suflScient  for  the  solution  of  all  problems  in  Simple  Percentage. 

EXAMPLES. 

1.  Bought  a  borse  for  $840,  and  sold  it  for  $560.  How 
much  did  I  lose  per  cent.  ?  Ans.,  33J%. 

SuG. — Here  h  and  s  are  given  to  find  r.  Hence  formula  (2)  is  to  be 
used.  And  as  there  is  loss  involved,  the  —  sign  is  to  be  taken.  Sub- 
stituting in  this  formula,  560  =:  — — — .     Solving  for  r  we  have 

5^  =  100-,.,  or?i°  =100  -  r;  whence  .=  100  -  ?55  =  "»=  33i. 
«-iO  '        3  3  3 

2.  A  n  amber  being  increased  by  2  equals  14.  Kequired 
the  increase  per  cent.  ?  Ans.,  16|%. 

Bug.— Formula  (1)  gives  2  =  — ;  and  (2)   gives  14  =  ttttt--- 

'Jrh 
From  which  we  are  to  find  r.    Multiplying  (1)  by  7, 14  =  — - .    Whence 

—  = ■ or  7r  =  100  4-r,  orr=  -7-  =165. 

100  100  ^   '  6 

3.  A  piece  of  cloth  sold  for  $779,  cash,  which  was  5%  off. 
Kequired  the  price  of  the  cloth.  Price,  $820. 

4.  Sold  40%"  of  my  wheat,  and  had  remaining  981  bushels. 
How  much  had  I  at  first  ?  Ans.,  1635  bushels. 

5.  A  man  sold  two  horses  at  $420  each  ;  for  one  he  re- 
ceived 25%  more,  and  for  the  other  25%  less  than  its 
value.     Required  his  loss.  Loss,  $56. 

SuG. — Letting  b  and  bi  be  the  values  of  the  horses  we  have  420=3 


316  BUSINESS    RULES. 

5(100-25)        ....      6,(1004-25)  „,        ..„  .       5^        ^ 

- — ,  and 420= -— .     .  • .  75ft  =  1255 1,  or  5  =  -5^  and 

100  J.UU  o 

o 

b-\-bi  =  -5i,  the  value  of  the  horses.     Now  5 1,  found  from  the  2nd, 
o 

o 

is  336,  and  as  -5i  —  840  =  the  loss,  we  have  896  —  840  =  56  =  the 
o 

loss. 

Those  who  are  not  famihar  with  algebraic  reasoning  will  prefer  to  find 
the  values  of  5  and  5i  from  the  two  formulas,  and  add  them  together. 

6.  A  man  sold  72  turkeys,  which  was  32%  of  the  number 
he  had  remaining.    How  many  had  he  at  first  ? 

Ans.,  297. 

7.  A  farmer  saved  annually  $145^,  which  was  33^%  of  his 
annual  income.    Required  his  income  ?      -4/i.s.,$436^. 

8.  A  merchant  having  400  barrels  of  cider,  sold  at  one 
time  45%  of  it;  at  another  time  20%  of  the  remainder. 
How  many  barrels  did  he  sell  in  all?    ^?is.,224  bbls. 

rh  45  X  400       ... 

OPEEATION.  P    =J^     =     —^^^  =  180. 

_r,6,  _    20  X  220  _ 
^'~I00"~  100       ~ 


...  p-\-p^=  224. 

9.    A  housekeeper  gave  to  her  neighbor  ^  of  a  pound  of 

tea,  and  had  f  of  a  pound  remaining.    What  per  cent. 

of  her  tea  had  she  remaining?  Ans.,  85f%. 

3         H     ^        7r  600        .^, 

OPEBATION.  -  =  _-,6=--     .-.r^  — =  85f 

10.  John  has  -|  of  a  dollar,  and  Henry  has  -|  of  a  dollar. 
What  per  cent,  of  John's  money  equals  Henry's? 
What  of  Henry's  equals  John's  ? 

Ans.,  120%,  83^%. 

r  =  120. 

[Note. — For  other  examples    in  percentage,  see    Stoddard's  Com- 
plete Arithmetic,  173-177  pp.] 


OPERATION. 

3  _  ^-i       6  -    ^ 
5        100'              20 

1  r|  3r 
2~"  100'               50 

SIMPLE  INTEREST,   AND  COMMON  DISCOUNT.  317 

SUCTION  11. 
Simple  Interest,  and  Common  Discount. 

[Note. — For  definitions  see  Elements  of  Arithmetic] 

8G,  I^rob,  1.  To  express  the  relation  between  principal, 
rate  per  cent,  time^  and  interest. 

Solution. — Letp,  r,  t,  and  i  represent  respectively  the  principal,  rate 
percent,  time  and  interest;  i  being  in  the  denomination  for  which  the 
rate  per  cent,  is  estimated.  Thus,  if  the  rate  per  cent,  is  rate  per  cent, 
per  year,  t  is  to  be  understood  as  years;  if  the  rate  per  cent,  is  per 
month,  i  is  months,  etc. 

ITD 

Then  as  p  is  the  base,  the  percentage  for  a  unit  of  time  is  ■—-  {84); 

and  for  t  units  of  time  it  is  -—•    .-.  i  =  — -—• 
lUU  100 

87*  I*rob,  2,  To  express  the  relation  between  amount, 
principal,  rate  per  cent,  and  time. 

Solution. — Since  the  amount  is  the  sum  of  principal  and  in- 
terest, representing  the  amount  by  a,  we  have  a  =  p  -{-  i.      But 

trp      „  ,     trp  100  4- fr  100a 

t  =  — -—'   Hence  a  =  r>  -4 i— ,    or  a  =  o ■ — •    .  •.  0= 

100  ^  ^  100  ^      100  -^     100+^r. 

88,  ScH. — This  problem  embraces  the  common  rule  for  Discount 
(see  Arithmetic,  268  et  sq.).  The  pupil  should  be  careful  to  under- 
stand the  reasonableness  of  discount.  For  example,  if  I  hold  a  note 
against  Mr.  B. ,  which  note  is  payable  at  any  future  time,  if  the  note  is 
drawing  interest  for  all  money  is  worth,  the  Present  Worth  of  the  note  at  the 
time  it  icas  given  was  its  face.  If  the  rate  at  which  it  is  drawing  interest 
is  less  than  money  is  worth  (or  if  it  draws  no  interest)  the  Present 
Worth  at  its  date  is  less  than  the  face  of  the  note.  (If  it  draws  no  in- 
terest, its  Present  Worth  at  any  time  is  less  than  its  face. )  Finally,  if 
the  rate  which  the  note  draws  is  greater  than  the  market  rate,  the 
Present  Worth  of  the  note  at  its  date  is  more  than  its  face. 


ScH. — The  two  formulce 

=  ^   and 
100 

100  +  tr 


(1)    i=^  and 


(2)    a=p-\-i=p- 


100 


318  BUSINESS   RULES. 

are  sufficient  to  solve  all  problems  in  Simple  Interest  and  Common  Dis- 
count. 

EXAMPLES. 

1.  What  is  the  interest  on  $250  for  1  yr.  10  mo.  15  da.,  at 
6% per  annum?  Ans.,  $28.12^. 

Solution. — In  this  example  I  am  to  consider  principal,  time,  rate 

per  cent,  and  interest,  the  latter  of  which  is  the  unknown  quantity. 

Formula  (1)  expresses  the  relation  between  these  quantities.     The  rate 

per  cent,  being  per  annum,  the  time  must  be  in  years.     15  days  =  ^g  = 

10  5 
.5  of  a  month.     10.5  months  =  — ^  :=.875  of  a  year.     .  • .  1  yr.  10  mo. 

3         5 

15  da.  =  1.875  years.    Now  (1)  gives  i  =  -^  = ^ = 

28.124. 

2.  What  is  the  interest  on  $47.25  for  1  tjr.  and  6  mo.,  at 

6%  per  annum  ?  Ans.,  $4.2525. 

4.721 
.3      3        -^A^ 

OPERATION.      I  = — — <^ =  4.2525. 

-20- 
4- 
-^■ 

3.  "What  is  the  interest  on  $145.50  for  1  yr.  9  mo.  24  da.,  at 
6%  per  annum  f  Ans.,  $15.86  nearly. 

.       1.816X6X145.50       ,^  __ 
OPERATION.     I  = =  15.86  nearly. 

4.  What  is  the  interest  on  $123.75  for  2  yr.  8  mo.  12  da. 
at  ()% per  annum?  Ans.,  $20.0475. 

5.  What  is  the  interest  on  $475  for  2  yr.  7  mo.  and  20  da., 
at  6%  per  annum  ?  Ans.,  $75.208|. 

6.  What  is  the  interest  on  $340.60  for  4  yr.  and  5  mo.,  at 
6%  per  annum  f  Ans.,  $90,259.     . 

7.  ¥/hat  is  the  interest  on  $50.40  for  1  yr.  and  10  mo.,  at 
7%  per  annum  f  Ans.,  $6,468. 


SIMPLE  INTEREST,  AND  COMMON  DISCOUNT.  319 

SxjG. — In  solving  this  example  the  operation  upon  paper  consists 
simply  in  multiplying  12.833  by  .504.  The  pupil  should  always  reduce 
the  written  details  of  his  arithmetical  work  to  the  minimum.  Thus, 
in  this  case,  he  sees  mentally  that  10  mo.  are  .  833  of  a  year.     Hence, 

^^00.40x1-833x7^  But  1.833  X  7  he  produces  mentaUy,  12.833 

100  504 
and  writes   12.833.     And  cancelling  the   100  from  50.40 

makes  it .  504.     Hence  the  written  work  should  only  be  as  f^/^p? 

in  the  margin.     In  practice,  nothing  but  this  multiphcation  ""^^"^ 

should  be  written  down.  6.4G7832 

Solve  the  following  by  thus  reducing  the  written  work  to 
a  minimum.  The  answers  are  given  as  in  practice  in  busi- 
ness. 

8.  "What  is  the  interest  on  $49.80  for  2  yr.  and  11  mo.,  at 
7%  per  annum  ?  Ans.,  $10.17. 

Opekation.     i  =  ^^-^^^J^^^'^^^^  =  2.9166  X  7  X  .498  =  10.17. 

9.  What  is  the  interest  on  $95.40  for  3  yj\  9  mo.,  at  8% 
per  annum  ? 

Opekation,    i  =  — '■ '- —  =  28.62.      The  operation  should 

be  performed  mentaUy.  Thus  8  X  3. 75  =  30.  Dropping  the  0,  and 
one  0  from  the  denominator,  and  for  the  10  remaining  in  the  denomi- 
nator removing  the  decimal  point  in  95.40  so  as  to  make  it  9.54,  we 
have  simply  to  multiply  9.54  by  3.  All  of  this  should  be  done  at  a 
glance,  without  writing  more  than  is  given  above. 

10.  What  is  the  interest  on  $196  for  5  yr.  7  mo.,  at  9%  per 
annum  ?  Ans.,  $98.49. 

11.  What  is  the  interest  on  $471.11  for  4  yr.  8  mo.,  at  1\% 
per  annum?  Ans.,  $164.89. 

12.  What  is  the  interest  on  $18.60  for  3  mo.  12  da.,  at  8% 
per  mo.  ?  Ans.,  $1.90. 

Opekation,  {  =  ?:l2ii^l2:^  =  1.7  x  3  X  .372  =  .372  X  5.1  = 
$L90. 

13.  What  is  the  interest  on  $400  for  150  days,  at  2h%  per 
month?  Ans.,  $50. 


320  BUSINESS   RULES. 

-       2  5 

400  X  5  X^ 
Opebation.     i  = ^  =  50. 

loe- 

14  What  is  the  interest  on  $1,000  for  2  mo.  12  da.,  at  1^% 
per  month  f  Ans.,  $36. 


15.  What  is  the  amount  of  $432.10  for  5  yr.  4  mo.  24  da.,  at 
7% per  annum  f  Ana.,  $595 .43. 

Opebaxiok.     a  =pl5^#:  ^432.10  ^2^^^^  =  4.321x137.8 
=  595.4338. 


•^'    lUU  ■"  100 


16.  What  is  the  amount  of  $325.25  for  2  yr.  9  mo.  12  da., 
at  6i%  per  annum  ?  Ans.,  $384.09. 

Opebation.     a  =  325.25^+4^^^21^  =.  384.09. 

100 


17.  In  what  time  will  $13,  at  6%  per  annum,  give  $0,975 
interest?  Ans.,  1  yr.   3  mo. 

SoiiUnoN. — Since  principal,  rate  per  cent.,  time  and  interest  are 

compared  i  — »■     gives  the  relation.     As  time  is  required,  I  solve  this 

equation  for  t,  and  have  i  = .     Substituting  the  given  values,  t  = 

rp 

.025 
50        .075- 

18.  In  what  time  will  $45.25,  at  6%  per  annum,  give  $1.81 
interest?  Ans.,  8  mo. 

19.  In   what   time    will  $70.50,    at  9%  per  annum,  give 
$31,725  interest?  Ans.,  5  yr. 

20.  In   what   time  "will  $140,    at   1%   per    annum,   give 
$10.861|-  interest?  Ans.,  1  yr.  1  mo.  9  da. 


SIMPLE   INTEKEST,   AND   COMMON   DISCOUNT.  321 

21.  In  what  time  will  $48.50,  at  G%  per  annum,  amount  to 
$56,187-4?  Ans.,  2  yr.  7  mo.  21  da. 

SuG. —Subtract  the  principal  from  the  amount  to  find    the    in- 
terest. 

22.  In  what  time  will  $248,  at  6%  per  annum,  amount  to 
$282,224  ?  Ans.,  2  yr.  3  mo.  18  da. 

23.  In  what  time  will  $700,  at  9%  per  annum,  amount  to 
$712.35?  Ans.,  2  mo.  10.5  +  da. 


24.  At  what  per  cent,  will  $325  produce  $3.25  interest  in  2 
months?  Ans.,  6%. 

trp 
Solution. — Same  as  above,  finding  r  from  the  equation  i  = -— ;. 

__  lOOt       100  X  3.25  _ 
**  "~     ip    ~    i  X  325     ~   • 

25.  At  what  per  cent.  wiU  $40  produce  $13.36  interest  in 
2  yr.  9  mo.  12  da.  ?  Ans.,  12%. 

26.  At  what  per  cent,  will  $125  produce  $32,375  interest 
in  3  ?/r.  6  mo.  ?  Ans.,  7|%. 

27.  At  what  per  cent,  will  $124  produce  $29.17 a  interest  in 
4  \jr.  3  mo.  10  da.'i  Ans.,  U%. 

28.  At  what  per  cent,  will  $2,360.25  amount  to  $2,470,395 
in  7  months  ?  Ans.,  8%. 

Suggestion. — Find  the  interest  and  then  proceed  as  before. 

29.  At  what  per  cent,  will  $230  amount  to  $249. 83f  in  11 
mo.  15  da.?  Ans.,  9%. 


30.    What  principal  will  in  3  yr.  8  mo.  15  da.,  at  6%  per 
awniwi,  give  $76,095  interest  ?  Ans.,  $342. 


322  BUSINESS    KULES. 


Solution. — Solving  i  =  —~  forp,  I  have 
lOOi       100X76.095       100  X  76.095  _ 


^  tr  6  X  3.7-iV  ^^-^5 

31.  What  principal  will  in  4  yr.  9  mo.  18  6?a.,  at  9%  pjr 
annum,  give  $65,016  interest?  J/i.s.,  $150.50. 

32.  "What  principal  will  in  8  y7\  8  mo.  12  cZa.,  at  5%  per 
annM?7i;  give  $147.9435  interest?  Ans.,   $340.10. 


33.  How  long  will  it  take  for  $200,  at  simple    interest,  at 
6%  per  annum,  to  amount  to  $500  ? 

Ans.,  25  years. 
Solution. — I  have  under  consideration  p,  r,  a,  and  i,  the  latter  of 
which   is   the  unknown  quantity.      The  relation  between   these  is 

a=p T-rp Solving  this  for  U  I  find 

_  100(a  —  V)_   100(500  —  200)  _  300  _ 
~         pr  ~  6  X  200  ~    12    ~      ' 

34.  How  long  will  it  take  $1,  at  simple  interest,  at  10% 
per  cent,  jjer  annum,  to  amount  to  $100  ? 

35.  How  long  will  it  take  $75,  at  interest  at  5%  per  an- 
num, to  amount  to  $100  ? 

80,  CoR. — To  find  the  time  required  for  a  principal  to 
double,  triple,  or  become  n  tim.es  itself  at  any  rate  of  simple  inter- 
est, ive  have  a  =  2p,  3p  or  np.     Hence  the  last  formula  becomes 

t  =  = for  the  time  required  to  double. 

pr  r     -^  ^ 

rp   ,  •  1         L        ,        100  X  2  .  ^  ^       ,     ^       100  X  3 

10  triple  we  have  t  =  ;  to  quadruple,  t  = ; 

to  become  n  times  itself  t  =  ■ -. 

ScH, — It  appears  that  the  time  in  such  a  case  is  independent  of  the 
principal,  as  might  have  been  anticipated. 

36.  How  long  will  it  take  $100  to  become  $500  at  S%  per 
annum  ?  Ans.,  50  years. 


SIMPLE    INTEREST,    AND    COMMON    DISCOUNT.       323 

37.  How  long  does  it  take  a  principal  to  double  at  6%  jjer 
annum  ;  at  7%  ;  at  5%? 

38.  Sold  property  amounting  to  $3,000,  on  a  credit  of  12 
mo.  without  interest.  Money  being  worth  8%  per 
annum,  what  sum  in  hand  is  equivalent  to  the  $3,000 
under  the  contract?   Verify  it.        Ans.,  %2,111.11  +. 

SoiiUTioN. — I  am  to  find  a  sum  which  put  at  interest  for  1  yr.  at  8% 
will  amount  to  $3,000.     I  therefore  have  given  a,  r,  and  t  to   find  p» 

The  relation  of  these  is  given  in  the  formula  a=  p  — -^ —   from 

100 

100a  ,.    ,^.  ,  300000       ^^^^  ^^    , 

which  p  =  -j^jjj-TT^  =  (m  this  case)  -^^^  =  2777. 77  +. 

39.  A  debt  of  $500  will  be  due  in  3  yrs.  without  interest. 
"What  is  its  present  worth  if  money  commands  6% 
per  annum '^  Ans.,  $423.73. 

40.  What  discount  should  be  allowed  for  the  present  pay- 
ment of  a  note  of  $400,  due  3  yrs.  5  mo.  hence,  the 
note  not  bearing  interest,  though  money  is  worth  6% 
per  annum  ?  Ans.,  $68.05. 

41.  A  man  buys  a  piece  of  property  to-day  for  $5,000, 
giving  his  note  with  security  at  12%  per  annum,  pay- 
able 2  yrs.  9  mo.  hence.  What  is  the  present  worth 
of  this  note  if  money  brings  in  market  6%  per 
annum  ?  Ans.,  $5,708.13  — . 

Operation.  a=p  ^^^^  =  50(100  +  12  X  U)  =  50  X  133  =, 
100«  665000         ^^^„  ^  _ 

««50-  P'  =m>  +  lF,  =  mOr  =  '">'■''  - 

42.  A  merchant  buys  $2,000  worth  of  goods  on  90  days 
time,  at  2%  a  month.  He  could  have  borrowed  money 
at  1^%  a  month.  How  many  more  goods  could  he 
have  bought  for  the  same  money  if  he  had  borrowed  ? 

Alls.,  $28.71  worth. 


324  BUSINESS    RULES. 

43.  July  20tli,  1869,  I  hold  a  note  for  $500  dated  April 
6th,  1867,  due  Jan.  1st,  1872,  and  bearing  interest  at 
1%  l^er  annum.  What  is  the  present  worth  of  this 
note,  money  being  worth  10% per  annum? 

Ans.,  $534.86+. 


*  00,  I*VOhm  To  find  what  each  payment  must  he  in  order 
'to  discharge  a  given  principal  and  interest  in  a  given  number 
of  equal  payments  at  equal  intervals  of  time. 

Solution. — Letp  represent  the  principal,  r  the  rate  per  cent.,  t  one 
of  the  equal  intervals  of  time,  n  the  number  of  payments,  (i.  e.,  nt  is 
the  whole  time),  and  x  one  of  the  payments. 

There  will  be  as  many  solutions  as  there  are  different  methods  of 
computing  interest  on  notes  upon  which  partial  payments  have  been 
made. 

1st.  By  the  United  States  Court  Rule. — As  the  payments  must  exceed 
the  interest  in  order  to  discharge  the  principal,  this  rule  requires  that 
we  find  the  amount  of  p,  for  time  t,  at  r  per  cent.     This  is  done  by 

multiplying  by  1  +  -—- ,  and  gives  p(  1  -f  -tht  )•    From  this  subtract- 

ing  the  payment  x,  the  new  principal  is  p(  1  -j-  --—  )  —  x.    Again  find- 

\  100  / 

ing  the  amount  of  this  for  another  period  of  time,  t,  and  subtracting 

the  second  payment  • 

In  like  manner,  after  the  third  payment  there  remains 
After  the  4th  payment,  the  remainder  is 
Finally,  after  the  ?ith  payment,  we  have 


SIMPLE   INTEREST, — PARTIAL   PAYMENTS.  325 

Whence 

'  i+o+w)+o+w)V(i+4y-o+4)""'' 

This  denominator  being  the  snm  of  a  geometrical  progression  whose 
first  term  is  1,  ratio  (1  -\- V  and  number  of  terms  n,  its  sum  is 


0  +  4)" 


Hence  x 
rt  /.    .     rt     ^ 

lUO  "      ■    ' 


2nd.    By  the  Vermont  Rule. — The  amount  of  the  principal  for  the 
whole  time  is  pfl-\-  ^ttt-  ]• 
The  amount  of  the  1st  payment  is a-l+   ttjttC^  —  1)    > 

-         -  "        2nd         "  ^[l+T50^'^-4 

"         "  "        3rd         -  a-[l  +    -^^{n  -  3)], 

Etc.,  etc.,         etc. 

((  ((  ____-__■•_         " 

The  nth  payment  (with  no  interest)  is x. 

The  sum  of  the  amounts  of  these  payments  is 

nx  +  ^x[(n-l)  +  (>i-2)  +  (n-  3) 1]. 

The  series  in  the  brackets  being  an  arithmetical  progression  whose 
first  term  is  (n —  1),  common  difference  —  1,  last  term  1,  and  number  of 

terms  (w  —  1),  its  sum  is  ( — - —  )n.  Hence  the  sum  of  the  payments  is 

nrt 

nic4--Yn77^( — o — )'^'  ^^'  ^L'^"' 9 J'    -^^^^  by  the  condition 

this  sum  equals  the  amount  of  the  principal;  consequently 


nrt 

-{11  —  !)• 


x  = . , 


326  BUSINESS    RULES. 

ScH. — If  the  payments   are   made   annually,  t  =  1.     And  letting 

7' 

r'  =  — -.   i.  e. ,  letting  the  rate  per  cent,  be  expressed  decimally,  the 
formulas  become 

pr'il  -j-  r')"     , 


By  the  U.  S.  Rule, 


(1  4-  r')''  -  1 
2p(l  4-  r'n) 


By  the  Vermont  Fade,       x  =  - 

zn-\^rwji  —  1) 

44.  AVhat  must  be  the  annual  payment  in  order  to  dis- 
charge a  note  of  $5,000,  bearing  interest  at  10%  per 
annum,  in  5  equal  payments  ? 

Ans.,  By  the  U.  S.  Rule,  $1,318.99  within  a  half  cent. 
By  the  Vermont  Rule,  $1,250. 

Query. — What  occasions  the  great  disparity  between  the  payments 
required  by  the  different  rules  ? 

45.  What  annual  payment  is  required  to  discharge  a  note 
of  $300,  bearing  interest  at  7  per  cent,  per  annum,  in 
4  equal  payments  ? 

46.  The  sum  of  $200  is  to  be  applied  in  jDart  towards  the 
payment  of  a  debt  of  $300,  and  in  part  to  paying  the 
interest,  at  6%  in  advance,  for  12  months,  on  the  re- 
mainder of  the  debt.  What  is  the  amount  of  the  pay- 
ment that  can  be  made  on  the  debt  ? 

SuG.  — Let  X  represent  the  payment;  then  (300  —  ic)  X  t'oo  is  the  in- 
terest on  the  remainder  of  the  debt;  and  we  have  therefore  the  equa- 
tion, X  +  (300  —  X)  X  T&o  =  200. 

Am.,  $193.62. 

47.  A  is  indebted  to  B  $1,000,  and  is  able  to  raise  but  $600. 
With  this  sum  A  proposes  to  j^ay  a  part  of  the  debt, 
and  the  mterest,  at  8%  in  advance,  on  his  note  at  2 
years  for  the  remainder.  For  what  sum  should  the 
note  be  drawn?  Ani^.,   $476.19. 


PARTNERSHIP.  'd'27 

SECTION    III. 

Partnership. 

[Note. — For  definitions,  see  Elements  of  Arithmetic] 

91.  JBrinclple.  In  Simple  Partnership,  i.e.,  ivhen  all 
the  capital  is  employed  the  same  length  of  time,  the  fundamental 
principle  is  that  the  shares  of  the  gains  shall  bear  the  same  ratio 
as  the  shares  of  the  stock. 

EXAMPLES. 

1.  A  and  B  enter  into  partnership.  A  puts  in  $1,200  and 
B  $1,800.  They  gain  $900.  What  is  each  one's  share 
of  the  gain  ? 


2  :  3. 


2.  A,  B,  and  C  enter  into  partnership.  A  puts  in  $340, 
B  $460,  and  C  $500.  They  gain  $390.  What  is  the 
gain  of  each  ? 

Ans.,  A's  $102,  B's  $138,  and  C's  $150. 

Opekation.  ic  +  Xi  -{-Xi^^  390.  x  :  ic,  :  x^  :  :  340  :  460  :  500,  or 
jc  _^  a-i  4-  X2  :  X  :  :  1300  :  340,  or  390  :  x  :  :  130  :  34,  or  3  :  x  :  :  1  :  34. 
.  • .  X  =  102.  X  +  Xi  4-  X2  :  Xi  : :  1300  :  460,  or  390  :  x,  ;  :  130  :  46. 
or  3  :  Xi  :  :  1  :  46.  .  • .  Xi  =  138.  x  -f-  x,  -f  x,  :  x^  :  :  1300  :  500,  or 
390  :  X2  :  :  130  :  50,  or  3  :  X2  :  :  1  :  50.     .  • .  x^  =  150. 

3.  Any  number  of  persons  as  A,  B,  C,  D,  etc.,  unite  in  a  part- 
nership, putting  in  respectively  a,  b,  c,  d,  etc.,  dollars 
each.    The  gain  or  loss  is  s.    What  portion  falls  to  each  ? 


Solution. - 

-Let 

X  and  Xi  be  their  respeeti 

Then 

x  +  Xi  =  900 

and 

X  :  Xi   :  :  1200  :  1800  . . 

Whence 

x  +  Xi    :    X  :  :  5  :  2 

or 

900          :    X  : :  5  :  2 

. 

•.  X  =  360,  andxi  =  540. 

328  BUSINESS   EULES. 

Solution. — Let  Xa,  Xh,  Xc,  xa,  etc.,  be  the  respective  shares  of  the  gaio 
or  loss.     Then  Xa -\-  Xb  -^  Xc  -\-  Xd  -\-,  etc.  —  s, 

.  and  Xa  :  Xb  :  Xc  :  Xd etc.  : :  a  :h  :  c  :  d etc. 

Whence  Xa  -\-  Xb  -{-  Xc  -\- Xd -\-  etc.  :Xa  :  ■  a  -\-b  -\-  c  -\-  d  -\-  etc.  :  a 
or    s  :  Xa  :  :  a  -\-  b  -^  c  -{-  d  -\-,  etc.  :  a.     ■ 

_  ^'^ 

"~a-f-6  +  c-f-t/-|-,  etc.* 

In  like  manner    Xb  = 


Xc 


a-\-b-\-c-^d-\-,  etc. 

cs 
a  -\-  b  -{-  c  -^  d  -{-,  etc' 

ds 

Xd  = . 

a-\-b-{-c-\-d-{-,  etc. 


ScH. — This  is  the  common  mle  for  Simple  Partnership,  viz.  :  Multi- 
ply the  gain  or  loss  by  each  partner's  stock  and  divide  the  products  by 
the  whole  stock. 

[Note, — In  such  examples  the  solution  is  so  simple  that  the  equa- 
tion scarcely  renders  any  assistance.  In  the  following  its  advantages 
will  appear.] 

4.  Two  men  commenced  trade  together.  The  first  put  in 
$40  more  than  the  second  ;  and  the  stock  of  the  first 
was  to  that  of  the  second  as  5  to  4.  What  was  the 
stock  of  each?  Ans.,  $200, and  $160. 

Opebation. — Let  x  =  what  the  first  put  in.  Then  x  —  40  =  what 
the  second  put  in.  And  we  have  x  :  x  —  40  :  :  5  :  4  or  x  :  40  :  :  5  :  1. 
.-.  x  =  200. 

5.  Three  men  trading  in  company  gained  $780,  which  was 
to  be  divided  in  proportion  to  their  stock.  A's  stock 
was  to  B's  as  2  to  3,  and  A's  to  C's  as  2  to  5.  What 
part  of  the  gain  should  each  have  received  ? 

Ans.,  A,  $156  ;  B,  $234 ;  C,  $390. 

6.  Three  men  trading  in  company,  put  in  money  in  the 
following  proportion  :  the  first,  3  dollars  as  often  as  the 
second  7,  and  the  third  5.  They  gain  $960.  What  is 
each  man's  share  of  the  gain  ? 

Ans.,  $192  ;  $448  ;  $320. 


PARTNERSHIP.  329 

7.  A,  B,  and  C  found  a  purse  of  money  ;  and  it  was  mu- 
tually agi'eed  that  A  should  receive  $15  less  than  one 
half,  that  B  should  have  $13  more  than  one  quarter,  and 
that  C  should  have  the  remainder,  which  was  $27.  How 
many  dollars  did  the  purse  contain  ?  A?is.,  $100. 


92*  J^rinciple.  In  Compound  Partnership ^  i.  q.,  part- 
nership in  which  the  several  partners'  shares  of  the  capital  are 
in  for  different  lengths  of  time,  the  gain  or  loss  is  divided  in 
the  ratio  of  the  products  of  the  several  amounts  of  stock  into 
the  times  which  they  respectively  remained  in  the  business. 
This  is  assuming  that  the  use  of  $a  for  time  t  in  business  is 
equal  to  $atyb?'  time  1. 

8.  A  and  B  enter  into  partnership.  A  furnishes  $240  for 
8  months,  and  B  $560  for  5  months.  They  lose  $118. 
How  much  does  each  man  lose  ? 

Ans.,  A  48,  and  B  $70. 

Solution. — Let  Xa  =  A's  share  of  the  loss,  and  Xb,  B's.  Then  Xa  -\- 
a;o  =  118,  and  x^  :  Xb  :  :8  X210  :5x  560  :  :  24  :35.  Whence Xa  +  X(,  :Xa  :  : 
59  :  24  or  118  :  iCa  :  :  59  :  24  or  2  :  iCa  :  :  2  :  48.  .  • .  Xa  =48,  and  Xu  = 
118  —  48  =  70. 

9.  A,  B,  and  C  entered  into  partnership.  A  put  in  $100 
for  4  months,  B  $300  for  2  months,  and  C  $500  for  3 
months.  They  gained  $250.  How  much  was  each 
man's  gain  ? 

Opebation.     Xa  -\-  Xb  -\-  Xc  =^  250,  and 

Xa  :  Xft  :  Xc  :  :  4  X  100  :  2  X  300  :  3  X  500  :  :  4  :  6  :  15.     Whence 
250  :  Xa  :  :  25  :  4,  or  10  :  Xa  :  :  1  :  4.     .  • .  Xa  =  40. 
250  :  X6  : :  25  :  6,  or  10  :  Xft  :  :  1  :  6.     .-.  Xb  =  60. 
250  :  Xc  :  :  25  :  15,  or  10  :  Xc  :  :  1  :  15.     .  • .  x^  =  150. 

10.  A,  B,  and  C  hire  a  pasture  for  $180.  A  puts  in  8  cows 
for  10  weeks,  B  20  for  5  weeks,  and  C  30  for  9  weeks. 
How  much  ought  each  to  ipaj  ? 

Ans.,  A  $32,  B  $40,  and  C  $108. 


330  BUSINESS   EULES. 

11.  To  gather  a  field  of  wheat,  A  furnished  8  laborers  for 
5  days,  B  12  for  3  days,  and  C  6  for  4  days.  For  the 
whole  work  A,  B,  and  C  received  $45.50.  How  much 
should  each  have  received  ? 

Ans.,  A  $18.20,  B  $16.88,  and  C  $10.92. 

12.  Any  number  of  persons,  as  A,  B,  C,  D,  etc.,  unite  in 
partnership,  A  putting  in  $a  for  time  ^  ;  B,  $6  for  time 
th ;  C,  $c  for  time  t, ;  D,  $d  for  time  t^,  etc.,  etc.  The 
gain  or  loss  is  s.  How  is  it  to  be  shared  by  the  part- 
ners? 

Solution. — Letting  Xa,  Xb,  Xc,  Xa,  etc.,  represent  the  respective  shares 
of  the  gain  or  loss,  we  have 

Xa  -\-  Xh  -^  Xc  -\-  Xd  -f-,  etc. ,  =  s,  and 
Xa  :Xb  :Xc  :xci etc.,  :  :  ata  :  hU  :  cle  :  dta etc.    Whence 

s:Xa   :  :  atu-\-Ut,-{-ctc-\-dtd  -  -etc.,  :ata.    .'.  Xa=  ,  ^^ s. 


s  :Xb  :  :  ata-{-htb-\-dc-]-dtd  -  -  etc.  :  Ub.    .' .   Xb=  ^^^   ,  ,^   ,  ^^   ,\,^ rrr*- 

And  in  Hke  manner  for  the  others. 


ala-\-btb-\-ctc-\-dtd-  -  etc, 

Ub 

ata-\-btb-\-ctc-\-dtd-  -  etc' 


ScH. — This  is  the  common  rule  for  Compound  Partnership,  viz.  : 
Take  the  produ.ct  of  each  partner's  share  of  the  stock  into  its  time  in 
trade.  For  any  partner's  share  of  the  gain  or  loss  multiply  the  whole 
gain  or  loss  by  the  product  of  his  stock  into  its  time,  and  divide  by 
the  sum  of  the  several  shares  of  the  stock  into  their  respective  times. 


SECTION  IV. 
Alligation. 

1.  A  farmer  mixes  together  10  bushels  of  oats,  at  40  cents 
a  bushel,  15  bushels  of  corn  at  50  cents  a  bushel,  and 
25  bushels  of  rye  at  70  cents  a  bushel?  What  is  the 
value  of  a  bushel  of  the  mixture  ?        Ans.,  58  cents. 


ALLIGATION.  331 

Solution. — Let  a  be  the  value  o\  a  bushel  of  the  mixture  of  which 

there  are  10  +  15  -f-  25  =  50  bushels.     Hence  50x  represents  the  value 

of  the  whole  mixture,  and  50a;  =  10  X  40  +  15  X  50  +  25  X  70  = 

2900.   .-.  x=:58. 

2.  A   grocer   mixes    120   pounds  of    sugar  at  5  cents  a 

pound,  150  pounds  at  6  cents,   and  130  pounds  at  10 

cents.     What  is  the  value  of  a  pound  of  the  mixtui'e  ? 

Ans.,  $0.07. 
8.  A  liquor  dealer  mixes  8  gallons  of  alcohol  100%,  12 
gallons  80%,  25  gallons  60%,  40  gallons  40%,  and  60 
gallons  20%  strong.     What  is  the  strength  of  the  mix- 
tui-e?  Ans.,  41fi%. 

4.  One  kind  of  wine  is  40  cents  a  quart,  and  another  24. 
How  much  of  each  must  be  taken  to  make  a  quart 
worth  28  cents  ? 
Statement,     x  -\-  y  =  1.     40x  -f  '^4?/  =  28. 


5.  Three  kinds  of  sugar  are  worth  respectively  6,  8,  and 
10  cents  a  pound.  How  much  must  be  taken  of  each 
to  make  a  mixture  worth  7  cents  a  pound  ? 

Solution. — Let  x,  y,  and  z  be  the  amounts  of  each  required,  a; 
pounds  at  6  cents  a  pound  are  6a:  ;  t/  at  8,  8?/  ;  z  at  10,  lOz.  The  whole 
amount  is  x  -\-  y  -{-  z,  and  the  whole  value  Q,x  -{-  ^y  -[•  lOz.  Hence 
6x  -f-  8?/  +  lOz  =  7(.x  -{-y  -\-  z).  But  here  are  three  unknown  quanti- 
ties, and  the  example  gives  but  one  set  of  conditions.  The  problem 
is  therefore  indeterminate  ;  that  is,  there  are  not  conditions  enough 
given  to  fix  the  values  of  the  unknown  quantities. 

Suppose  we  add  the  two  conditions  :  To  make  a  mixture  of  48  lbs. ; 
and  that  twice  as  much  of  the  6  cent  sugar  shall  be  used  as  of  both 
the  others.  We  then  have  the  two  additional  equations  x  -\-y  -\-  z  =» 
48,  and  x  =2(y  -\-  z).  These  equations,  together  vrith  the  former, 
6x  -f  8?/  4-  lOz  =  7[x-\-y  -\-  z)  readily  give  x  =  32,  y=8,  and  z  =  8. 

93.  ScH. — The  last  example  is  a  case  in  Alligaiion  Alternate,  as  it  is , 
denominated  in  our  Arithmetics.  Such  examples,  as  they  are  usually 
stated,  are  simply  problems  in  which  there  are  more  unknown  quan- 
tities than  equations,  and  are  hence  Indeterminate.     The  Indetermi- 


332  BUSINESS  BULES. 

nate  Analysis  can  not  be  treated  in  this  volume,  but  a  few  further  illus- 
trations of  such  examples  will  be  given.  All  the  various  methods  of 
treating  this  subject  usually  given  in  our  Arithmetics  are  exceedingly 
cumbrous  and  perplexing  to  the  pupil,  and,  after  all,  fail  to  give  a  full 
view  of  it.  The  propriety  of  puzzhng  pupils  with  any  of  them  is 
exceedingly  questionable.  They  are  very  clumsy  and  incomplete 
efforts  at  doing  a  thing  which  becomes  very  simple  when  the  proper 
principles  are  developed,  which  principles  cannot  be  brought  forward 
in  Common  Arithmetic. 

6.  How  mucli  of  each  sort  of  grain,  at  48, 50,  and  68  cents 
a  bushel,  must  be  mixed  together,  so  that  the  com- 
pound will  be  worth  60  cents  a  bushel  ? 

Statement,  x,  y,  and  z  being  the  amounts  of  each  kind  respective- 
ly, we  have  48a;  +  50y  -f  68z  =  60(ic  -\-  y  -\-  z)  ov  &£  -\-  5y  =  4z. 
Now  any  real,  positive  values  may  be  assigned  to  either  two  of  these 
unknown  quantities,  which  will  give  a  positive  value  for  the  other. 

Thus  if  z  =  4,  and  y  =  2,  6x  =  16  —  10,  or  x  =  1.  Verify  these  re- 
sults. 

Again,  if  z  =  5,  and  y  =  1,  6x  =  20  —  5,  or  x  =  2^.  Verify  these 
results. 

Again,  if  z  =  6,  and  t/  =  3,  6x  =  24  —  15,  or  x  =  1^.  Verify  these 
results. 

In  like  maner  an  unlimited  number  of  sets  of  answers  can  be  ob- 
tained. 

J£,  however,  we  try  z  =  2,  and  y  =  3,  we  have  x  =  —  1^;  The  neg- 
ative sign  in  this  case  shows  an  impossibility.  The  cause  of  this  is 
evident  when  we  notice  that  2  bushels,  at  68  cents,  and  3  at  50,  make  5 
bushels,  worth  286  cents,  or  57i  cents  per  bushel.  This  mixture  can- 
not be  made  worth  60  cents  per  bushel  by  patting  in  grain  worth  only 
48  cents. 

The  —  1^  of  the  48  cent  grain  indicated  by  this  result,  may  be  un- 
derstood to  mean  that  we  are  to  take  out  of  the  5  bushels,  worth  286 
cents,  1^  bushel,  worth  48  cents  per  bushel.  This  leaves  3|  bushels, 
worth  230  cents,  or  just  60  cents  per  bushel. 

7.  A  merchant  has  two  kinds  of  wine.  The  first  kind  is 
worth  12  shillings  per  gallon,  and  the  second  is  worth 
7  shillings  per  gallon.  How  many  gallons  of  each  kind 
must  he  use  in  order  to  form  a  mixture  worth  9  shil- 
lings per  gallon  ? 


ALLIGATION.  333 

Alls.,  Letting  x  and  y  represent  the  quantities  re- 
spectively, he  may  take  any  quantity  he  pleases  of 
either,  so  that  he  takes  such  an  amount  of  the  other  as 
to  preserve  the  relation  3x  ==  2y.  That  is,  he  must 
take  1-^  times  as  much  of  the  latter  as  of  the  former. 
If  he  takes  2  of  the  former,  he  must  take  3  of  the  lat- 
ter. If  he  takes  6  of  the  former,  he  must  take  9  of 
the  latter,  etc.,  etc. 

8.  How  much  corn  at  48  cents,  barley  at  36  cents,  and 
oats  at  24  cents  per  bushel,  must  be  taken  to  make  a 
compound  worth  30  cents  per  bushel  ? 

SuG. — Show  that  he  can  take  any  amount  he  pleases  of  either  one, 
if  he  takes  proper  amounts  of  the  other  two  respectively.  Show  what 
he  may  take  of  the  second  and  third  kinds  if  he  take  1  bushel  of  the 
first.  Is  there  any  limit  to  the  number  of  ways  he  can  make  up  the 
mixture  when  he  takes  1  bushel  of  the  first  kind  ?  Can  he  take  3  bush- 
els of  the  first  kind  and  2  of  the  second  ?  What  value  would  this  give 
to  z  (representing  the  amount  of  the  third  kind)  ? 

9.  A  merchant  wishes  to  mix  32  pounds  of  tea  at  36  cents 
per  po^ind,  with  some  at  48  cents,  and  some  at  72 
cents.  How  many  pounds  of  each  kind  must  he  take 
to  form  a  mixture  worth  56  cents  per  pound  ? 

SuG. — The  relation  is  2?/  —  x  =  80.  Any  value  for  either  x  or  y  may 
be  taken  which  gives  a  positive  value  for  the  other;  and  any  positive 
value  of  X  will  give  in  this  case  a  positive  value  for  the  y,  for  2^  =i  80  -|-  a;. 
Kx  =  4,  i/  =  42.     If  ic  =  10,  y  =  45.     lix  =  \,y  =  40|^,  etc. ,  etc. 

If  we  add  to  this  example  the  condition  that  the  whole  amount  of 
the  mixture  shall  be  102  pounds,  the  problem  becomes  determinate  ; 
as  there  are  then  two  equations  with  two  unknown  quantities.  The 
equations,  when  reduced,  are  2?/  —  x  =  80,  and  y  -{•  x  =  70.  "Whence 
jc  =  20,  and  y  =  50. 

10.  A  man  bought  horses  at  .$50  each,  oxen  at  $40,  cows 
at  $25,  calves  at  $10,  so  that  the  average  price  per 
head  was  $30.     How  many  were  bought  of  each  ? 

SuG. — Of  course  fractional  values  as  well  as  negative  are  excluded 
by  the  nature  of  this  example.  The  conditions  to  be  met  are  4.C  + 
2y  =  z  -f-  4io.  negative  and  fractional  values  being  excluded  by  the 
nature  of  the  problem.  Can  the  conditions  be  met  by  taking  3  cowa 
and  5  calves  ?  Why  ?  Can  they  by  taking  2  cows  and  5  calves  ?  How  ? 
Can  they  by  taking  4  horses  and  6  cows  V    How? 


334  BUSINESS   RULES. 

11.  A  bought  240  barrels  of  molasses  for  $4,320 ;  worth, 
respectively,  $10,  $14,  $20,  and  $22  ;  how  many  barrels 
of  each  did  he  buy?  Ans.,  40,  20,  160,  and  20. 

SuG.  — The  conditions  are  x-\-y-\-z-\-w=  240,  and  4:X-{-2y:=z-{-  2w, 
negative  and  fractional  values  being  excluded  by  the  nature  of  the  exam- 
ple. Here  are  two  equations  with  four  unknown  quantities,  hence  any 
other  two  conditions  maybe  imposed  which  do  not  conflict  with  the  na- 
ture of  the  example.  Can  x  =  GO  and  ?/  =  20  ?  Yes,  This  reduces  the 
equations  to  z-\-w  =  160,  and  z-j-  2m5  =  280;  from  which  to  =  120,  and 
z  =  40.  Can  the  condition  that  half  the  quantity  shall  be  of  the  first 
two  kinds  be  met  ?  No.  This  gives  z  -\-  w  =  120,  which  subtracted 
from  z  -^  2io  =  4:X  -{-  2y,  makes  lo  =  2(2iK  -f-  y)  —  120.  But  since 
x-{-y  =  120,  2  (2.x  -{-y)—  120  >  120,  .  • .  ro  >  120,  which  would  make 
z  negative. 

12.  I  have  two  kinds  of  molasses  which  cost  me  20  and  30 
cents  per  gallon  ;  I  wish  to  fill  a  hogshead,  that  will 
hold  80  gallons,  with  these  two  kinds.  How  much  of 
each  kind  must  be  taken,  that  I  may  sell  a  gallon  of 
the  mixture  at  25  cents  per  gallon  and  make  10  per  cent, 
on  my  purchase  ? 

Ans.,  58-^\  of  20  cents,  and  21^\  of  30  cents. 

13.  A  lumber  merchant  has  several  qualities  of  boards  ;  and 
it  is  required  to  ascertain  how  many,  at  $10  and  $15  per 
thousand  feet,  each,  shall  be  sold  on  an  order  for  60 
thousand  feet,  that  the  price  for  both  qualities  shall  be 
$12  per  thousand  feet. 

Ans.,  36  thousand  at  $10,  and  24  thousand  at  $15. 

14.  How  many  ounces  of  gold  23  carats  fi.ne,  and  how  many 
20  carats  fine,  must  be  compouuclod  with  8  ounces  18 
carats  fine,  that  the  alloy  of  tb'j  three  diiferent  quah- 
ties  may  be  22  carats  fine  ? 

Ans.,  48  oz.  of  the  first,  and  8  oz.  of  the  second. 

[Note, — These  applications  might  be  extended  to  much  greater 
length,  did  space  permit.  The  equation  renders  important  aid  in  many 
problems  in  Compound  Interest,  but  their  discussion  usually  requires 
a  knowledge  of  Quadratics,  and  some  of  them  ot*  liogaritnms.  Thejf 
must,  therefore,  be  reserved  for  the  future.  ] 


PURE   QUADRATICS.  335 

OHAPTEE  lY. 

QUADBATIC   EQUATIONS, 


sucTiojsr  I, 

Pure  Quadratics. 

94,  A  Quadratic  Equation  is  an  Equation  of 
the  second  degree  {0,  8). 

95,  Quadratic  Equations  are  distinguished  as  Pure 
(called also //icompfefe),  o^ndi  Affected  (called  also  Complete.) 

96,  A  I^ure  Quadratic  Equation  is  an  equation 
which  contains  no  power  of  the  unknown  quantity  but  the 
second  ;  as  a^^  _j_  ^  :=  ^^^  ^2  —  2>h=^  102. 

97,  An  Affected   Quadratic  Equation  is  an 

equation  which  contains  terms  of  the  second  degree  and 
also  of  the  first,  with  respect  to  the  unknown  quantity  or 
quantities ;  as  x'^ —  4^  =  12,  hxy — x — y^=  16a,  imxy  -\-y:x=b. 

98,  A  Moot  of  an  equation  is  a  quantity  which  sub- 
stituted for  the  unknown  quantity  satisfies  the  equation. 


99,  I*roh,  To  solve  a  Pure  Quadratic  Equation. 

R  ULE. — Transpose  all  the  terms  containing  the  unknown 

QUANTITY  INTO  THE  FIRST  MEMBER,  AND  UNITE  THEM  INTO  ONE, 
CLEARING  OF  FRACTIONS  IF  NECESSARY.  TRANSPOSE  THE  KNOWN 
TERMS  INTO  THE  SECOND  MEMBER.  DiVIDE  BY  THE  COEFFICIENT 
OF  THE  UNKNOWN  QUANTITY.  FiNALLY,  EXTRACT  THE  SQUARE  ROOT 
OF  BOTH  MEMBERS. 

Dem. — According  to  the  definition  of  a  Pure  Quadratic,  all  the  terms 
C9ntaining  the  unknown  quantity  contain  its  square.     Hence  they  can 


336  QUADRATIC    EQUATIONS. 

bo  transposed  and  united  into  one  by  adding  with  reference  to  tlie 
Bquare  of  the  unknown  quantity.  That  transposition,  and  division  of 
both  members  by  the  same  quantity  do  not  destroy  the  equahty  has 
aheady  been  proved.  [Let  the  student  repeat  the  reasoning.  ]  Ex- 
tracting the  square  root  of  the  first  member  gives  the  first  power  of 
the  unknown  quantity,  i.  e.  the  quantity  itself.  And  taking  the  square 
root  of  both  members  does  not  destroy  the  equation,  since  like  roots  of 
equal  quantities  are  equal. 

100,  Cor.  1. — Every  Pure  Quadratic  Equation  has  two  roots 
numerically  equal  but  loith  02J2^osite  signs.  For  every  such 
equation,  as  the  process  of  solution  shows,  can  be  reduced 
to  the  form  x^^=a  {a  representing  any  quantity  whatever). 
Whence,  extracting  the  root,  we  have  x  =  -\-  \/a  ;  as  the 
square  root  of  a  quantity  is  both  +,  and  —  (203,) 

ScH. — The  question  naturally  arises,  Why  not  put  the  ambiguous 
sign  (the  -f )  before  the  x,  as  well  as  before  the  second  member  ?  It  is 
proper  to;  but  there  is  no  advantage  gained  by  it.  Thus,  if  we  write 
4-  icz=  4-  s/a,  we  have  -|-  a;=  +  \/a,  or  — ic  =+  Va.  But  the  former 
is  ic  =  4-  Va,  and  the  latter  becomes  so  by  changing  the  signs  of  both 
members.  So  that  all  we  learn  in  either  case,  is  that  x  =-{-  \/a,  and 
X  =  —  \/a. 

101,  Cor.  2. — The  roots  of  a  Pure  Quadratic  Equation  may 
both  be  imaginary,  arul  both  will  be  if  one  is.  For  if  after 
having  transposed  and  reduced  to  the  form  x'^=^a,  the  second 
member  is  negative,  as  ^^  :::=  —  a,  extracting  the  square  root 
gives  x  =  -^  V —  a,  and  x  =  —  v —  a,  both  imaginary. 

EXAMPLES. 

1.  Given  8^2  —  lo  —  x^  =  12  +  4tx^  —  54  to  find  the  vahie 
of  X. 

MODEL  SOLUTION. 

Operation.     3x2  —  10  —  x^  =  12  -^  4x2  —  54, 

(2)  —  2x2  =  —  32. 

(3)  x2  =  16, 

X  =  -I-  4. 


PURE    QUADRATICS.  337 

Explanation.  — Transposing  and  uniting  terms,  I  ha.ve  —  2x-  =  —  32. 
[K  necessarj%  let  the  pupil  show  why  the  equality  is  not  destroyed.  ] 
Dividing  both  members  by  —  2,  I  have  x^  =  16.  Extracting  the  square 
root  of  both  members,  I  find  x  =  4.  4  (read,  ' '  x  equals  plus  and  minus 
4").  [Be  careful  and  not  omit  showing  why  the  equation  is  not  de- 
stroyed by  the  processes,  as  long  as  there  is  a  possibility  of  a  doubt 
that  it  is  imderstood.  ] 

Veetfication. — Substituting  -f-  4  for  x,  the  equation  becomes  48  — 
10  — 16  =  12  -f-  64  —  54,  or  22  =  22.  Substituting  —  4  gives  just  the 
same  since  —  4  squared,  ( —  4)2,  is  the  same  as  -}-  4  squared. 

2.  Given  x-^  -{- 1  = -— -{- 4:,  to  find  x.  Hoots,  x  =  ^2. 

4z  — 

_    ^.  x(9-^2x)       3^-f  6  .     .    , 

6.  Criven    -— = ,  to  nnd  x. 

10  5 

Boots,  a:  =  4-  3. 

4.  Given [- =  -,  to  find  the  values  of  x. 

4:-\-   X  4: X  6 

Boots,  a:  =  +  1. 

5.  Given  x'^  —  ah  =  d,  to  find  the  values  of  x. 

Boots,  X  =  -\^  vd  +  ab. 

6.  Given 3  +  t^  =-  777  —  ^^  +  ^A    to   find    the 

values  of  x.  Boots,  x  =  4-3. 

7.  Given  13  —  v'dx-^  -f  16  =  5,  to  find  the  values  of  x. 

Boots,  0*  =  -4-  4. 
2a 


8.  Given  x  -f  \/x-^  -f-  a  =  — "  to  find  the  values  of  x. 

V  x-^a 

Boots,  X  =  -^  I V  8a. 

SuG. — Clear  of  fractions.     Transpose  and  condense.     Square  both 
members. 

9.  Given   —^==  =  a:  +  ^x^-{-d.         Boots,  x=  ±2. 

v/a:2-f  5 

10.  Given    \/x  +  a=-^  Boots,  x  =  \/2a{a-{-b)+bK 

v^x — a 


'6'6t 

i 

QUADRATIC   EQUATIONS. 

11. 

Given    -  - 

a 

a 

X 

^1  —  a-^ 

12. 

Given  —  - 

X 

-  X 

2^77111       n  -^    ,                 ,   ,—        /- 

X               X 

13. 

Given 

a-          ^  a^ 

Boot^,  X  =  4-  V  a2  —  52, 

14.  Given  12  —  ;r^  :  ^x^  : :  100  :  25.  Boots,  x=±2. 

OPEEATION.     12  —  ic2  :  ^x'^  :  :  100  :  25 

12  —  ic2  :  a;2  :  :  50  :  25  :  :  2  :  1 
12  :  x2  :  :  3  :  1 
4:x2:  :1  :1    .-.  x=4:  2. 

[Note. — Use  the  principles  of  proportion  in  solving  these.] 

15.  Given  ^x^-  +  ij;^  —  3  :  ^x^  —  ^x^  -]- 3  :  :  9  :  3. 

Boots,  X  =  ^  4v^2. 

16.  Given  ^{x'^  —  5)^  :  j;^  —  5  :  :  2  :  1.        Boots,  ^=  +  3. 

OPERATION.     ^(x2  —  5)2  :  x2  —  5  :  :  2  :  1 
x2  —  5  :  1  :  :  4  :  1 
x2  -  5  :  5  :  :  4  :  5 
x2  :  5  :  :  9  :  5    .  • .  x  =  +  3. 

17.  Given  ^(11  +  ^')  :  1(4^^  —  2)  :  :  5  :  2. 

Boots,  ;r=  4-  2. 

18.  Given  x^  + 'i  :  x:^  —  11  :  :  100  :  40.  _ 

Boots.  x=^  +  v/21. 

19.  Given      ^      ,      =2 — .        Boots,  x=  +  2. 

^3  +  4  x^  —  4  — 

/r4-4       r 4        10 

20.  Given  — ^-  -f =■  =  ^.  Boots,  x=  +  8. 

X — 4       x-^4c         6  — 

21.  Given  (x  +  2)2  =  Ax  +  5.  Boots,  x=±l. 

22.  Given  -^(.r^  _  12)  =  i:r'  —  1.  Boots,  x==  ±6. 

x+1  x—1  7 


23.   Given 


072  —  7^       ^2  _[_  7^7       072  —  73* 

BootSf  X 


PURE    QUADRATICS.  339 


24.  Given  x  -\-  ^ x'-  —  ^2  —  ix  =  1.  Eooi^,  x ■■ 


±h 


Query. — If  the  members  are  squared  as  the  equation  stands  will  it 
be  simplified?  (See  Abt.  26,  and  examples  under  it  for  hints  as  to 
methods  of  freeing  of  radicals. ) 


25.  Given  Va  +  x  =  '^ x+  ^'x- 


Boots,  x-^^  -^  v/flj  —  b-. 


26.  Given    ^1  -\-  x-^  +   ^^  1  +  ^^  +  v^l  —  ^-  =  v^l  —  x'-. 
Verify.  Boots,  ar==  +  fv  — 6. 

27.  Given   Vb^  +  x^  =^  \/a-'  +  x'-. 


Boots,  x=  ±  ^  v^{b'  —  a^). 


28.  Given  h  —  v/a^  +  x'^ 


a-  —  x^ 


a  +  Va-^  _|-  x-^ 

Boots,  x=  4-  -/wM •• 

—  a — 6  N  a 

__    ^.        \/a''-\-x"~a       ,  _    ^  2av/6 

29.  Given :=  h.  Boots,  a;  =  + -r 

V  a-^ -{- x^ -{- a  J- — ^ 


30.  Given      ^'^'+^  +  2 ^ ^^  ^^^^^^  x  =  -^^. 

\/3j7^  +  4— 2  "" 

31.  Given  -===  ^.  Boots,  =^^^5- 

VX-^+1  +  VJ7--2  — 1 

32.  Given 1 -— -— -  =  x. 


33.  Given 


^  H-  V  2  —  x-^      X  —  v^2  —  X- 

Boots,  a;  =  -f  ^. 
1  la 


Boots,  j:  =  +  ^a^. 

34.  Given ==-.  i?ooifs,  ^=  +         _\ 

V  a-  +  .r-  —  jr      ^  -*  v^ftc 


34C 

1 

QUADRATIC 

EQUATIONS. 

35. 

Given 

1              + 

1 

1 

v/1- 

-x^-^1      VI 

—  j;2- 

_1           ^'* 

iioo^s,  a:  = 

=  +iv/3. 

36. 

Given 

1  + 

1+37 

—  n  ■     ■ 

1  — 

X 

X  -{-  ^l+x-^ 

1  —  ^  + 

^1  +  x-^ 

Boots,  X 

=  ±^{a~ 

-2y-l. 

37. 

Given 

'       + 

i 

=  ax. 

J7  +  V  2  —  x-^      X  —  v/2  —  x^ 

Moots,  X  =  -\-   j^l . 

—    y     a 


38.  Given  — ^= +    , =  a. 


Boots, 


4    I         4 
—  a\  a^ 


APPLICATIONS. 

1.  What  two  numbers  are  those  whose  «um  is  to  the  great- 
er as  10  :  7,  and  whose  sum  multipHed  by  the  less  pro- 
duces 270?  yl/is.,  21  and  9. 

SuG. — Let  lOx  =  the  sum  of  the  numbers,  and  7x  the  greater. 

ScH.  — It  is  customary  to  omit  the  negative  roots  in  giving  answers 
to  examples,  the  nature  of  which  renders  such  answers  impossible. 
In  this  case  the  question  is  about  pure  number,  and  hence  the  answers 
should  be  given  without  signs. 

2.  There  are  two  numbers  whose  ratio  is  that  of  4  to  5, 
and  the  difference  of  whose  squares  is  81.  What  are 
the  numbers  ?  Ans.,  12  and  15. 

3.  What  two  numbers  are  those  whose  difference  is  to  the 
greater  as  2  to  9,  and  the  difference  of  whose  squares 
is  128  ?     Verify. 


PURE    QUADRATICS.  341 

4.  Find  three  numbers  which  bear  the  same  ratio  to  each 
other  as  ^,  f ,  and  f  do  to  each  other,  and  the  sum  of 
whose  squares  is  724,  Numbers,  12,  16,  18. 

5.  Find  three  numbers  in  the  ratio  of  m,  n,  and  p,  the 
sum  of  whose  squares  is  equal  to  a. 

„     7                  I     am'                   I      an-' 
Numbers,    +k\ ; — ,    +v >    and 


4. 


ap2 


6.  Divide  14  into  two  psirts  so  that  the  greater  part  di- 
vided by  the  less  shall  be  to  the  less  divided  by  the 
greater  as  16  to  9- 

SuG. — Having  '■ : :  :  16  :  9,   it    follows    that    x*  : 

^14   —  X  X 

(14  —  a;)2  : :  16  :  9,  and  ic  :  14  —  ic  : :  4  :  3,  and*  ;  14  : :  4  :  7.     .  •.  x  = 
8,  and  14  —  a;  =  6. 

7.  Divide  a  into  two  parts  so  that  the  greater  part  divid- 
ed by  the  less  shall  be  to  the  less  divided  by  the 
greater  as  m  to  n. 

„     ,         a^^m-  ,       ttv^n 

Parts,  -—z — —,  and  — = — — . 

ScH. — Example  7  is  example  6  generalized.     The  pupil  should  de- 
duce the  results  in  the  former  from  thesa     Thus,- substituting  «  =  14 

m  =  16,  and  n  =  9,  — — — :  =  8,  etc. 

8.  What  two  numbers  are  they,  whose  product  is  126,  and 

the  quotient  of  the  greater  divided  by  the  less,  3^  ? 
Generalize  this.  Ans.,  6  and  21. 

9.  The  sum  of  the  squares  of  two  numbers  is  370,  and 
the  difference  of  their  squares  208.  Required  the 
numbers.  Numbers,  9  and  17. 

SuG. — If  X  represents  the  first,  370  —  x-  is  the  square  of  the  second 
number. 


342  QUADRATIC   EQUATIONS. 


10.  Generalize  the  9th,  and  show  that  ^\/2(8  +  d)  and 
-^x/2(s  —  d)  are  general  results. 

J.1.  For  comparatively  small  distances  above  the  earth's 
surface  the  distances  through  which  bodies  fall  under 
the  influence  of  gravity  are  as  the  squares  of  the  times. 
Thus,  if  one  body  is  falling  2  seconds  and  another  3, 
the  distances  fallen  through  are  as  4  :  9.  A  body  falls 
4  times  as  far  in  2  seconds  as  in  1,  and  9  times  as  far 
in  3  seconds.  These  facts  are  learned  both  by  obser- 
vation and  theoretically.  It  is  also  observed  that  a 
body  falls  IB^^-  feet  in  1  second.  How  long  is  a  body 
in  falling  500  feet?  One  mile  (5280  ft. )  ?  Five  miles  ? 
Ans.,  To  fall  500  ft.  requires  5.58  seconds.  To  fall 
5  miles  requires  40.51  seconds. 

1 1.  A  and  B  lay  out  some  money  in  a  speculation.  A  dis- 
poses of  his  bargain  for  $11,  and  gains  as  much  per 
cent,  as  B  lays  out.  B  succeeds  in  gaining  $86  ;  and 
it  appears  that  A  gains  four  times  as  much  per  cent, 
as  B.     Eequired  the  capital  of  each. 

Results,  $5  =  A's  capital,  and  $120  ==  B's. 

SuG.— To  avoid  fractions  let  4«  represent  B's  capital,   and  conse- 
quently A's  gain  per  cent. 

13.  A  money  safe  contains  a  certain  number  of  drawers. 
In  each  drawer  there  are  as  many  divisions  as  there 
are  drawers,  and  in  each  division  there  are  four  times 
as  many  dollars  as  there  are  drawers.  The  whole  sum 
in  the  safe  is  $5,324  ;  what  is  the  number  of  drawers  ? 

Ans.,  11. 
SuG. — This  example  gives  rise  to  a  cubic  equation,  4x3  ^^  5324. 

14.  Two  travelers,  A  and  B,  set  out  to  meet  each  other  ; 
A  leaving  the  town  C  at  the  same  time  that  B  left  D. 
They  travelled  the  direct  road  from  C  to  D,  and  on 
meeting  it  appeared  that  A  had  travelled  18  miles  more 


PUKE    QUADKATICS.  343 

than  B  ;  and  that  A  could  have  gone  B's  journey  in 
15f  days,  but  B  would  have  been  28  days  in  perform- 
ing A's  journey.  What  is  the  distance  between  C 
andD?  Ans.,  126  miles. 


SUG. 


M  D 

J 1 


I                                                                      I  I 

A        • >  < —  B 


If  X  :=  CM  =  the  distance  A  travelled,  then  x  —  18  =  MD  =  the  dis- 

a; ig  3. 

tance  B  travelled.      -   ■.-,—    =  distance  A  travelled  a  day  ;  and  -^ 
lof  ''  28 

=  distance  B  travelled  a  day.     N'otice  that  the  times  are  equal. 

15.    From  two  places  at  an  unknown  distance,  two  bodies, 

A  and  B,  move  toward  each  other  till  they  meet,  A  going 

a  miles  more  than  B.      A  would  have  described  B's 

distance  in  n  hours,  and  B  would  have  described  A's 

distance  in  m  hours.     What  was  the  distance   of  the 

two  places  from  each  other  ?  \/^  -f-  \/n 

Ans.,  a  X  -7=- T^. 

vm  —  V  n 

16.  A  and  B  engaged  to  work  for  a  certain  number  of 
days.  A  lost  4  days  of  the  time  and  received  $18.75. 
B  lost  7  days  and  received  $12.  Now  had  A  lost  7 
and  B  4  days,  the  amounts  received  would  have  been 
equal.  How  long  did  they  engage  to  work  and  at 
what  rates?  Ans.,  Whole  time,  19  days. 

Sua. — If  X  =  the  whole  time,  what  represents  A's  daily  wages? 
What  B's  ?  Alter  the  equation  is  formed,  see  if  you  cannot  strike  out 
a  numerical  factor  from  both  members,  and  extract  the  root  without 
expanding. 

17.  A  vintner  drew  a  certain  quantity  of  wine  out  of  a  full 
vessel  that  held  256  gallons  ;  and  then  filled  the  vessel 
with  water,  and  drew  off  the  same  number  of  gal- 
lons as  before,  and  so  on  for  four  draughts,  when  there 
were  only  81  gallons  of  pure  wine  left.  How  much 
wine  did  he  draw  each  time  ? 

Ans.,  64,  48,  36,  and  27  gallons. 


344  QUADRATIC  EQUATIONS. 

SuG. — If  he  drew  out  -  part  of  the  contents  of  the  cask  each  time, 

X 1 

there  remained  after  the  first  drawing? th  of  the    wine  :  after  the 

X 

second    X or ,   and    after  the  fourth 


X  X  X-  x^ 

(x  —  l)^         _,  x  —  1 

•••  -^r-=-^V..or— -=^. 

18.  A  number  a  is  diminished  by  the  nth  part  of  itself, 
this  remainder  is  diminished  by  the  ntli  part  of  itself, 
and  so  on  to  the  fourth  remainder,  which  is  equal  to 
b.     Eequired  the  value  of  n.  y  ~ 

19.  There  is  a  number  such  that,  if  the  square  root  of  three 
times  its  square  +  4,  be  taken,  the  quotient  of  this 
root  increased  by  2,  divided  by  the  root  diminished  by 
2,  is  3.     What  is  the  number  ? 

QuEKY. — Which  of  the  equations  in  the  preceding  part  of  this  sec- 
tion does  this  give  rise  to  ? 

20.  If  the  square  root  of  the  difference  between  the  square 
of  a  certain  number  and  2,  be  both  added  to  and  sub- 
tracted from  the  number  itself,  the  sum  of  the  recip- 
rocals of  the  result  is  j'-  of  the  number  itself.  What 
is  the  number  ? 

Query. — Which  of  the  equations  in  the  preceding  part  of  this  sec- 
tion does  this  give  rise  to?     With  what  modifications? 


SECTION  IL 

Affected  Quadratics. 


102.  An  Affected  Quadratic  equation  is  an 
equation  which  contains  terms  of  the  second  degree  and 
also  of   the  first  with  respect  to  the  unknown  quantity. 


AFFECTED    QUADRATICS.  345 

.^  .        «            2ax4-Sbx-^       ^a-^x^ 
a:2  —  3j:  =  12, 4  j:  +  ^ax'^  = ,  and ^ax  +  36'^  =  0 

are  affected  quadratic  equations. 

103.  Proh,  To  solve  an  Affected  Quadratic  Equaiion. 

RULE. — 1.  Keduce  the  equation  to  the  form  x--\-ax^=h, 
the  characteristics  of  which  are,  that  the  first  member 
consists  of  two  terms,  the  first  of  which  is  positive  and 
simply  the  square  of  the  unknown  quantity,  its  coefficient 
being  unity,  while  the  second  has  the  first  power  of  the 
unknown  quantity,  with  any  coefficient  (fl)  positive  or  nega- 
tive, integral  or  fractional  ;  and  the  second  member  con- 
sists of  known  terms  (6). 

2.  Add  the  square  of  half  the  coefficient  of  the  second 
term  to  both  members  of  the  equation. 

3.  Extract  the  square  root  of  each  member,  thus  produc- 
ing A  SIMPLE  equation  FROM  WHICH  THE  VALUE  OF  THE  UNKNOWN 
QUANTITY  IS  FOUND  BY  SIMPLE  TRANSPOSITION. 

Dem. — By  definition  an  Affected  Quadratic  Equation  contains  but 
tliree  kinds  of  terms,  viz. :  terms  containing  the  square  of  the  unknown 
quantity,  terms  containing  the  first  power  of  the  unknown  quantity,  and 
known  terms.  Hence  each  of  the  three  kinds  of  terms  may.  by  clearing 
of  fractions,  transposition,  and  uniting,  as  the  particular  example  may 
require,  be  united  into  one,  and  the  results  arranged  in  the  order 
given.  If,  then,  the  first  term,  L  e.  the  one  containing  the  square  of 
the  unknown  quantity,  has  a  coefficient  other  than  unity,  or  is  nega- 
tive, its  coefficient  can  be  rendered  unity  or  positive  without  destroy- 
ing the  equation  by  dividing  both  the  members  by  whatever  coefficient 
this  term  may  chance  to  have  after  the  first  reductions.     The  equation 

will  then  take  the  form  x-  ±(ix  =  ±  h.     Now  adding  (-)   to  the  first 

member  makes  it  a  perfect  square  ( the  square  of  x  -f  -  j,  since  a  trino- 
mial is  a  perfect  square  when  one  of  its  terms  (the  middle  one,  ax,  in 
this  case)  is  +  twice  the  product  of  the  square  roots  of  the  other  two. 
these  two  being  both  positive  (123 f  Pakt  I. )  But,  if  we  add  the  square 
of  half  the  coefficient  of  the  second  term  to  the  first  member  to  make 


346  QUADRATIC  EQUATIONS. 

it  a  complete  square,  we  must  add  it  to  the  second  member  to  preserve 
the  equality  of  the  members.  Having  extracted  the  square  root  of 
each  member,  these  roots  are  equal,  since  like  roots  of  equals  are  equal. 
Now,  since  the  first  term  of  the  trinomial  square  is  x-,  and  the  last 


(I) 


does  not  contain  x,  its  square  root  is  a  binomial  consisting  of  x  4- 


the  square  root  of  its  third  term,  or  half  the  coefl&cient  of  the  middle 
term,  and  hence  a  known  quantity.  The  square  root  of  the  second 
member  can  be  taken  exactly,  approximately,  or  indicated,  as  the  case 
may  be.  Finally,  as  the  first  term  of  this  resulting  equation  is  simply 
the  unknown  quantity,  its  value  is  found  by  transposing  the  second 
term. 

EXAMPLES. 

1.  Given  6  —  x tt  ="  ^  —  2^+— -to  find  the  value  of 

o  0 

Xj  and  verify. 

MODEL    SOLUTION— OPERATION. 

fi  —  x—^'  —  a-  —  2-1-  4-  — 

48  —8x—  x2  =  8x  —  18  -J-  x2 
—  2x2  _  i6x  =  —  66 

x2  -f-  8x  =  33 
xi  -f  8x-|-16=33  +  16=49 

X  +  4  =  +  7, 
x=  +  7  —  4=3,  and  —  11. 

Explanation. — Clearing  the  equation  of  fractions,  transposing  and 
uniting,  and  dividing  both  members  by  —  2,  I  have  x^  -|-  8x  =  33. 
[Give  the  reasons  why  these  processes  do  not  destroy  the  equation 
often  enough  to  keep  them  fresh  in  memory.  ]  Now  since  8x  contains 
the  square  root  of  x-  as  one  of  its  factors,  the  other  factor,  8,  is  twice 
the  square  root  of  the  other  term  of  a  trinomial  square  {123 f  Pakt  I). 
Hence  ^  of  8,  or  4,  squared,  16,  is  the  third  term.  Adding  this  term  I 
have  x2  _j_  8x  -f- 16,  which  is  a  perfect  square.  But  as  I  have  added  16 
to  the  first  member  to  make  it  a  perfect  square,  I  must  add  it  to  the 
second  member  to  preserve  the  equality.  This  gives  x^  -f  8x  -f  16  = 
49.  Extracting  the  square  root  of  both  members,  I  have  x  -j-  ^  = 
-f  7.  [Notice  that  (x  -f-  4)-'  =  x^  -|-  8x  +  16,  and  tell  why  this  does 
not  destroy  the  equation.  ]  Finally,  transposing  the  4,  I  have  x  = 
—  4  4-  7,  t.  c,  X  =:  3,  and  —  11,  3  if  I  take  the  square  root  of  49  as  -f 
7,  and  —  11  if  I  take  the  square  root  of  49  as  —  7.  Both  are  correct. 
Hencn  there  are  two  roots  (values  of  or)  of  this  equation,  3,  and  —  11. 


AFFECTED    QUADRATICS.  347 

Verification. — To  verify  the  value  x  =  3,  I  substitute  3  for  a;  in  the 
given  equation,  and  have  6  —  3  —  t  =  3  —  2:^-|-8»oJ^  3  —  |  =  f-f- 
I,  or  ^f  =  -^if-.  To  verify  the  value  x  = — ^11, 1  substitute  for  x,  —  11, 
in  the  given  equation  and  have  6  -f-  H  —  "^1^  =  —  H  —  2^  -f-  ''"a^.  or 
17  —  -HJ^  =  -  13i  -f  H^  or  ^i-  =  \^. 

2.  Given  x-^  —  8.r  +  5  =  14,  to  find  the  values  of  x,  and 
verify.  Result,  j;  =  9,  and  —  1. 

3.  Find  the  roots  of  the  equation  x-  —  12^  +  30  =  3, 
and  verify.  Roots,  9  and  3. 

4j; 9 

4.  Find  the  roots  of  x  —  2  =  — . 


SuG.  — This  reduces  to  x^  —  6x  =  —  9.  Whence,  completing  the 
square,  x"^  —  6x  +  9  =  9  —  9  =  0,  and  x  —  3  =  0,  or  x  =  3.  In  this 
case  it  appears  that  the  equation  has  but  one  root. 

5.  Given  -  =  — '■ to  find  the  vahies  of  x. 

5  X 

6.  Find  the  roots  of  S{x  —  4)  = ^- —. 


Sua— This  reduces  to  x'^  —  8x  =  —  20.  Whence,  completing 
the  square  x-  —  8x  -)-16  =  16  —  20  ==—-  4,  and  extracting  the  root 
X  4  =   -f  2^/"!,  or  X  =  4  4-  2^^ —  1,  two  imaginary  roots. 


7.  Find  the  values  of  x  m 


5       X  —  5  * 


Results,  a;  =--  5  +  2^—5,  and  5  —  2^—5. 

300       „„        o  ^  —  300 

8.  Find  the  roots  oi  x  -] -f  73  =  3  . 

X  X 

Roots,  —  30,  and  —  40. 

„4(2j^  +  6)         8 

9.  Find  the  roots  of  - — ^r = ^. 

2x  X 

Only  one  root,  —  2. 


348  QUADBATIC    EQUATIONS. 

10.  Find  the  roots  of  l{x  +1)  +  Slf-Z 1  =  Q, 

Boots,  5(—  1  +  v/— 1),  tncl  5(—  1  —  v/_l),  or  as 
the  same  may  be  written,  —  5(1  —  v^ — 1)^  and 

—  5(1  + v'^T). 

11.  Find  the  roots  of  |^-^  —  ^.x  +  20^  =  42f,  and  verify. 

Boots,  7,  and  —  6^. 
ScH.  1. — This  process  of  adding  the  square  of  half  the  coefficient 
of  the  first  power  of  the  unknown  quantity  to  the  first  member,  in 
order  to  make  it  a  perfect  square,  is  called  Completing  the  Squake. 
There  are  a  variety  of  other  ways  of  completing  the  square  of  an 
affected  quadratic,  some  of  which  will  be  given  as  we  proceed  ;  but 
this  is  the  most  important.  This  method  will  solve  all  cases  :  others 
are  mere  matters  of  convenience,  in  special  cases. 

12.  Given  x^  —  ^  +  3  =  45  to  find  its  roots,  and  verify. 

13.  Given  2x^  +  8^  —  20  =  70  to  find  its  roots,  and  verify. 
IOj;        _,  5.^2  —  65 

and  verify. 

35 3j. 

15.  Given  Gx  + =  4:4:  to  find  the  values  of  x. 

X 

SuG. — Clearing  of  fractions,  Qx^  4-35  —  3x  =  44a;, 
Transposing  and  uniting,  6x^  —  47a;  =  —  35, 
Dividing  by  6,  x^  —  ^^x  =  —  ^^, 

Completing  the  square,  x^  —  ^^x  +  (H)^  =  (ff  )^  —  Y  =  ^W, 
Extracting  the  root,  x  —  f |-  =  4.  f f. 
Transposing,  x  =  ^  ±  f  J  =  7,  or  f . 

16.  Find  the  roots  of  5x '■ =  2x  -\ '—- . 

X  —  3  2 

SuG. — Notice  the  compound  negative  term.      Cleared  of  fractions 
and  reduced,  the  equation  becomes  x^  —  3a;  =  4.     .  • .  x  =  4,  and  —  1» 

17    -c.-    ^   ^1.  .       4^  16  100—9^ 

17.  Fmd  the  roots  of =  3. 

X  4^2 

Suggestion. — Multiply  by  4x-  to  clear  of  fractions. 


14.  Given  5 ^  =  13^^ '■ to  find  its  roots 


AFFECTED  QUADKATICS.  349 

18.  Find  the  roots  of  3x^  —  20j:  —  62  =  Ix  —  2x^  +  100. 

Boots,  9,  and  —  3f . 

19.  Find  the  roots  of  2a;  —  2  =  2  +  -. 

X 

Boots,  3,  and  —  1. 

20.  Find  the  roots  of  -^^^  —  ^a;  +  7  =  8  —  |. 

Boots,  f ,  and  —  f. 

48 

21.  Find  the  values  of   x  in  the  equation h  5  = 

^  ^  +3 

165  ^      , 

— — TTT-  Besult,  X  =  b%  and  5. 

8j7         20 

22.  Find  the  values  of  x  in  the  equation  — — =6. 

x-\-'A        "dx 

Besult,  X  ==  10,  and  —  f . 

23.  Given  ^{x  +  4)  —  l-I— ^  =  i  (4^  +  7)  _  l,  to  find 

X    O 

the  values  of  x.  Besult,  x  =  21,  and  5. 

24.  Given  ^ 1 — —  =  5^-^,  to  find  the  roots  and 

X  .r  +  12 


verify. 

Find  t] 
and  verify. 


25.    Find  the  roots  of  ^(8  —  x) =  U^  —  2), 


oa    T?-  A  ^i.  ^      «  2x  +  9   ,   4^—3  3:r  —  16 

26.  Find  the  roots  of  — - —  +  - — --  =  3  + , 

y  4:X  -\-  6  10 

and  verify. 

27.  Find  the  values  of  x  in  the  equation  Sx^  —  2ax  =  b. 


Besult,  X  =  — = — ~ ■ 


350  QUADRATIC    EQUATIONS. 

OPERATION.  3a;2  —  2ax  =  6; 

Dividing  by  3,       cc'^ —x  =  -, 

^        1  ..       .1                     ,        2«       ,    /ay-         a2         5       a2_j_3& 
Completing  the  square,  x^ —x  +  (o)    =n    +0=  — z ^ 

Extracting  the  square  root,  a;  —  —  =  4-  -v^a-  -\-  '6b, 

^  .  a       "/ a!^  -\-  oh  «  4-  '^«'  +  <^^ 

Transposing,  a;  =  —  4- »    or  — = 

00  o 

ScH. — The   form  — ^ ; — ~ —  signifies  that  there  are  two  values 

o 

of  ic  ;  i.  e.,  that  each  of  the  signs  4"  and  —  i^iay  be  used.     Thus  the 
Values  in  this  case  are 

a  -J-  >/a2  +  36         ,          or.  —  ^/as -(-  36 
— '   and  X  = -; 


3 


*28.  Find  the  roots  of  3x'^  -\-  hax  =  m. 


—  5a  -i-  ^25a-  +   12m 

Boots, ■ — =^ -^ — 

b 

29.  Find  the  roots  of  da'^b^x^  —  Ga^b'-x  =  bK 

a   4-  v/a2  _^  52 


Boots, 


3a'6-^ 


X      a       2 
30.  Find  the  values  of  x  in  the  equation  ■-  -\ —  =  -. 

a       X       a 


31.  Find  the  roots  of 


BesuU,  X  =  1  -j-  v^l  —  a-\ 

2x(a  —  x)  a 

'6a  —  2x  4* 

Boots,  fa,  and  ^a. 


104,  CoR.  1. — An  affected  quadratic  equation  has  two  roots. 
These  roots  may  both  be  positive,  both  be  negative,  or  one  posi- 
tive and  the  other  negative.  They  are  both  real,  or  both 
imaginary. 

Dem. — Let  X-  -j-px  =  q  be  any  affected  quadratic  equation  reduced 
to  the  form  for  completing  the  square.  In  this  form  p  and  q  may  be 
either  positive  or  negative,  integral  or  fractional.     Solving  this  equa- 


AFFECTED   QUADRATICS.  351 


tion  we  have  x  =  —  ^  +  kJj, — I"  9-  ^^  ^^^^  ^^^  observe  what  dif- 
ferent forms  this  expression  can  take,  depending  upon  the  signs  and 
relative  vahies  of  p  and  q. 

1st.    Wheyi  p  and  q  are  both  positive.     The  signs  will  then  stand  as 

given;  i.  e.,  x  =  —  -■  ±  \/  "T"  +  5-  Now,  it  is  evident  that 
.1^ — h  5  >  "T"'  ^^^  \l ^"  '?  ^^  ^^®  square  root  of  something 

p2  p  I  p"  p  p- 

than  -^.    Fence,  as  —  <  ^—  -{- q,  —  ^   +   \Jx  "^  ^^     ^^ 


more 


positive;  but  —  ^  —  \/~T — h  ?  i^  negative,  for  both  parts  are  nega- 
tive. Moreover  the  negative  root  is  numerically  greater  than  the  posi- 
tive, since  the  former  is  the  numerical  sum  of  the  two  parts,  and  the 
latter  the  numerical  difference.  .*.  When  p  and  q  are  both  -|-  in  the 
given  form,  one  root  is  positive  and  the  other  negative,  and  the  nega- 
tive root  is  numerically  greater  than  the  positive  one.  See  Example  1, 
above. 

2nd.    When  p   is   negative  and  q  positive.    We    then  have   x=  — 

— ^  ±      I  — T ^  5  =  7  ±      I      4-  ^-     I^  "^^  take  the  plus  sign  of 

the  radical,  x  is  positive;  but  if  we  take  the  —  sign,  x  is  negative,  since 


J 


1  "I"  ?  ^  o-     Moreover,  the  positive  root  is  numerically  the  greater. 

Wh'en  p  is  negative  and  q  positive,  one  root  is  positive  and  the  other 
negative  ;  but  the  positive  root  is  numerically  greater  than  the  nega- 
tive-    See  Example  2,  above. 

—  P 
8rd.    When  p  and  q  are  hoth  negative.     We  then  have  x  :=^ ^r^  4- 


real,  and  as  it  is  less  than  — ,  both  values  are  positive.     See  Example  3. 


li  J  =:q,      \—  —  g=0  and  there  is  but  one  value  of  x,  and  this  is 

positive.     (It  is  customary  to  call  this  two  equal  positive  roots,  for  the 
sake  of  analogy,  and  for  other  reasons  which  cannot  now  be  appreoi- 


352  QUADRATIC   EQUATIONS. 


-f<^.J 


ated  by  the  pupil.)     See  Examples  4  and  5.     I^  i"  <C  ^>  ^  |x  —  ?  ^®- 

comes  the  square  root  of  a  negative  quantity  and  hence  imaginary. 
See  Examples  6  and  7. 

4th.    When  p  is  positive    and  q  negative.     We   then  have  x  z=  -^ 

~  -1-       ' q.      As  before,  this  gives  two   real  roots  when  q  <C^' 

When  this  is  the  case  both  roots  are  negative.     [Let  the  pupil  show  how 

this  is  seen.  ]     When  q  =—,  the  roots  are  equal  and  negative  ;  i.  e.  there 

is  but  one.  When ^<^q  both  roots  are  imaginary.  See  Examples 
8,  9,  and  10. 

[Note. — It  is  not  important  that  the  pupil  remember  all  these  forms  ; 
but  it  is  an  excellent  exercise  to  give  the  discussion.  The  ingenious 
student  can  put  the  results  in  a  very  neat  analytical  table.  ] 

ScH. — It  may  be  asked  why,  when  we  extract  the  square  root  of  each 

r)2  pi 

member  of  the  equation  x"^  -\-  px  -\-  ^  =^  q  -\-  -j,  ^&  WTite  the  am- 
biguous sign  only  before  the  root  of  the  second  member.  The  reason 
is  the  same  as  given  under  Pure  Quadratics,  Art.  100,  Scholium.  Thus, 
in  strict  propriety,  the  square  root  of  each  member  of  this  equation 

being  taken  or  indicated  gives  +  fx-{---\=  ^      1^  _j_  g.     But  take 

these  signs  in  any  order  we  can,  it  amounts  to  taking  the  roots  as  hav- 
ing like  signs  (both  -{-,  or  both  — )  or  unlike  signs  (one  -f-  and  the 
other  — ).  Hence  it  is  sufficient  to  give  the  ambiguous  sign  to  one 
member  only  ;  and  it  is  most  convenient  to  give  it  to  the  second. 


32.  Given   v/;r  +  5  X  ^^  +  12  =  12,    to    find    the    values 
of  X.  Result,  x=  4:,  and  —  21. 

Suggestion. — First  clear  the  equation  of  radicals. 


33.  Given  v  4^  +  5  X   vlx  +  1  =  30,    to   find   the   values 
of  X.  Values,  x  =  5,  and  —  6^. 


34.  Given  x-{-5=  v  ^  _|_  5  -|-  (]^  to  find  the  values  of  x. 

Values,  X  ==  4l,  and  —  1. 


35.  Given  ^  +  16  —  7^a;+  16  =  10  — 4^^  +  16,  to  find 
the  values  of  ec.  Values,  j;  =  9,  and  — 12. 


AFFECTED    QUADRATICS.  353 


SuG. — Put  the  equation  in  the  form  x  -{-  6  =  Ss/x  +  16,  and  then 
square. 


36.  Given  S^x  +  6  +  2  =  a;  +  v'a;  +  6,  to   find   the   values 
of  X.  Result  J   a;  =  10,  and  —  % 

1         4 

37.  Given  v  H —  =  — =  to  find  the  roots. 

Resulty   x=  \^,  ^^. 

4 
Si3a. — Multiply  by  y  and  transpose,  and  y^ -y  =  —  1.       Com- 

4  4       4  1 

pletingthe  square,  y'^ ::  y-j-  q  =^q  —  ■'^^^q-     Extracting  the  root, 

y ^=  + v/i.     .-.  y==-^4-  71  =  2^1+  Vl=^s/l,   andvi 

v/3       ~    _  v^3  ~ 

=  s/3,  andi\/3. 

lOo,  CoR.  2. — ^/i  affected  quadratic  being  reduced  to  the 
form  x2  +  px  =  q,  the  value  of  x  can  always  he  written  out 
without  taking  the  intermediate  steps  of  adding  the  square  of 
half  the  coefficient  of  the  second  term,  extracting  the  root,  and 
transposing.  The  root  in  such  a  case  is  half  the  coefficient  of 
the  second  term  taken  ivith  the  oj^jDOsite  sign,  -f  the  square  root 
of  the  sum  of  the  square  of  this  half  coefficient,  and  the  known 
term  of  the  equation.     This  is  observed  directly  from  thefoimn 

X  =  —  9  ±  \/ V  ~i"  ^1'  ^'^^  'niore  in  detail  in  the  demonstration 
of  the  preceding  corollary. 

[Note. — The  pupil  should  use  this  method  in  practice,  but  be  care- 
ful that  the  complete  method  and  its  fuU  demonstration  is  not  lost 
Bight  of.  ] 

38.  Write  out  the  roots  of  the  following  without  going 
through  the  operations  of  completing  the  square,  etc.  : 
22  4.  4^;  =  60  ;  y'  —  4y  =  60  ;  x'^  +  16^7  =  —  60  ; 
x-2  _  16a;  =  —  60. 

39 .  Reduce  the  following  to  the  form  x'^  -f  px  =  q,  and 
then  write  out  the  roots  as  above  : 


354:  QUADRATIC   EQUATIONS. 

3^2  -f  2,r  4-  6  =  11,  gives  x^  +  ^x  =  f .     Whence  x  = 
—  i  +  ^i  +  i  =  —  ^  ±  t  =  1.  and  —  f . 
9j;2       30j;  ,     .  20  64      ^,,, 

Yq-\-  "^~  ="  ~~  ^'  ^^^^^  ^'  +  "3"*^  ^  ~  ~^'    ^^®^^® 


_       10  llOO        64  _  —  10  -f  6 

"^-"T  ±N   9~""y"~  3 


=—  =  —  14,  and  • 


-5J. 


4:X —  =  14,  gives  x^  —  t^  =  7-     Whence  x  = 

X  +  1  ^  4 


9         l81 
8±N64 


9  +  23 
+  1=  -^—  =  4,  and  —  If. 


40.  Given  dx~^  —  12^  ^  =  —  3,  to  find  the  values  of  x. 

Values,  X  =  3,  and  1. 


9 
Suggestion,     dx-'^  =  — •• 


y_  I y_ 

y 1  — 

y  y 


41.  Given    ^f-  +    f  =  ^^-,  to  find  the  roots. 


SuG. — Reduce  the  complex  fractious  to  simple  oues  by  multiplying 

numerator  and  denominator  by  v.      Whence     ■ 1-  '     '"     z=  -i-i 

Multiply  by  4(t/2  _  1)  and  4r/2  4.  4  -f-  %^  -f  8i/  +  4  =  13y^  —  13. 
.  • ,  1/  =  3,  and  —  |. 

42.  Given  va  -]-  x  -\-  va  —  x  ^ — ,  to  find  x. 

bV a  -\-  X 

Result,    X  =  -—  and  -— . 
5  5 

43.  Given  vx  -{-a  —  V^x  -}~  b  =  V%x,  to  find  the  roots. 


Result,  X  =  — ■  - — -- —  +  \^  2a-  +  2^2. 
'A 

SuQ.  x/x4-a  —  v'2a;  =  -/x  4-6.  x  -\-  a  ~  ^y/'lx^  -\-  2ax  4-  2x 
=  a;  +  6.  —  2 V'2x2~+~2a^  =  6  —  a  —  2x.  8x2  4- 8ax  =  &2  _  2a& 
—  4hx  4-  a^  4-  4ax  4-  4x2. 


AFFECTED   QUADRATICS.  355 

44.  Given     ^  ^\  —  {a^—b^)x= —^ ■ to 

find  the  values  of  x.  HesuU,    x  z=i  a,  and  —  b, 

SuG.     ; Y  =  "X" — T-     Whence  the  given  equation 

becomes  x-  —  (a  —  b)x  =  ab. 

X  ,  X  b 

45.  Given  —^ 7=^=   +    "7^ 7^=^^   ""  -7=.  to 

V  ^  +  V  a  —  ^  v.r —  va  —  x  vx 

64.   \/f)-  —  2aJb 
find  the  roots.  lioots,  x  =  — — = ; >• 


2xV 
)ns  in  the  first  member,  and 

Whence  2x^  =  26x  —  ab 


SuG. — Add  the  fractions  in  the  first  member,  and ■  =     /-• 

2x  —  a       V  X 


FOR  REVIEW  OR  ADVANCED  COURSE. 

100,  Cor.  3. —  UjDon  the  principle  that  the  middle  term  of  a 
trinomial  square  is  twice  the  product  of  the  square  roots  of  the 
other  two  {94^  OS,  Part  l.),we  can  often  complete  the  square 
more  advantageously  than  by  the  regular  rule. 

[Note.— A  few  examples  will  be  given  to  illustrate  this  corollary, 
and  if  more  are  needed  the  pupil  can  use  those  which  precede.  The 
methods  which  follow  are  rather  such  as  experts  will  use  than  such  as 
are  important  in  an  elementary  course  of  training.  The  use  of  them 
might  be  left  optional  with  the  student.  Those  who  have  special  ap- 
titude for  the  science  will  learn  them  with  profit,  while  to  perplex 
others  with  them  may  be  an  injury  rather  than  a  benefit.  ] 

46.  Solve  4^2  4.  \Qx  =  33. 

Solution. — Dividing  16a;  by  twice  the  square  root  of  4x2,  i.  e.,  by  4x, 
and  adding  the  square  of  the  quotient,  (4)-,  to  both  members,  4x2  -f. 
16x  +  16  =  49.  Extracting  the  root  2x  -f  4  =  +  7.  .  • .  x  =  §, 
and  — J-/. 

47.  Solve  8^2— 12a:  =  36. 

SuG. — Divide  the  equation  by  2,  and  proceed  as  above.  4x2  —  6x-f- 
(1)2  =  fii,  2x  —  f  —  4-  I,  and  x  =  3,  and  — l^-.  In  this  example  the 
regular  method  is  better. 


356  QUADRATIC    EQUATIONS. 

48.  Solve  3^2  ^  2x  =  5. 

SuG.  —Multiply  by  3  and  9x2  -f  6a;  =  15.  Whence  9x2  -f-  6x  -{- 1  =  16, 
x=l,  and  —  f . 

49.  Solve  110^2  —  21^  =  —  1.        Besidt,  x  =  ■^^,  and  -Jy- 

SuG.— Multiply  by  110,  and  (110)2a;2  — 21  •  110x  =  — 110.  Whence 
1102x2  —  21  .  110x-)-(^/)2=  i,  and  llOx  —  V=  ±  h- 

50.  Solve  3^2  ^  5^  =  2. 

SuG.  (3)2x2  4-  3  .  5x  +  (I )2  =  6  +  \i  =  Aa.  Whence  3x  +  f  =  +  f . 
.  • .  X  =  I,  and  —  2. 

ScH It  appears  that  by  this  method  the  term  to  be  added  to  com- 
plete the  square  is  the  square  of  i  the  coefficient  of  the  first  power  of  .r. 
Therefore  when  this  coefficient  is  odd,  fractions  arise.  These  can 
always  be  avoided  by  doubling  the  equation  when  this  coefficient  is 
odd,  before  completing  the  square. 

51.  Solve  3:r2  _  7^  =  40. 

Solution. — Multiplying  by  2  to  avoid  fractions,  6x2  —  14x  =  80. 
(6)2x2  —  6 .  lix  +  (7)2  =  49  +  480  =  529.  6x  —  7  =  ±  23.  x  =  5,  and 
-2|. 

52.  Solve  ^x{x  +  5)  =  ^6  +  4:x{l  —  x),  and  verify. 

SuG. — Perform  as  few  multiplications  as  possible.  When  the  square 
is  completed,  this  stands  (14)2x2  +  14  •  22x  -f-  (11)2  =  14  .  192  -|-  121 

=  2809. 

53.  Solve  4-^2  4-  i^  -f  y7_3_  =  0. 

1  2  S 

54.  Solve  - 


x  —  2       X  +  2       5 
55.  Solve  {ax  —  b)(bx  —  a)  =  c\ 

SuG.  a&x2  —  (a2_f_&-2)a;  =  c2  —  ah.  2abx^—{m)  =■=  2c2—  2ah,  letting 
(m)  stand  for  the  term  which  becomes  the  middle  term  of  the  trino- 
mial square  and  which  disappears  in  the  subsequent  process.  Then 
(2a6)2x2  —  (m)  +  (a2  -(-  62)2  ^  (^2  4.  j^iyz  _j_  4a6c2  —  4a262,      .  • .  x  = 

a2  -j-  &2  4.  v/(a-  — 6^)-4-4a6c2 


56.  Solve  ,  =  — '■ — .        Boots,  x  =  a,  and  la. 

a  +  V2ax  —  x-^      «— ^ 

Suggestion. — Clear  of  fractions  and  condense. 


EQUATIONS   SOLVED   AS   QUADBATICS.  357 


SECTION   III, 

Equations  of  Other  Degrees  which  may  be  solved  as 
Quadratics. 

107,  JProp,  1,  Any  Pure  Equation  {i.e.,  one  containing 
the  unknown  quantity  affected  with  hut  one  exponent)  can  he 
solved  in  a  manner  similar  to  a  Pure  Quadratic. 

Dem.  — In  any  such  equation  we  can  find  the  value  of  the  unknown 
quantity  affected  by  its  exponent,  as  if  it  were  a  simple  equation.  If 
then  the  unknown  quantity  is  affected  with  a  positive  integral  expo- 
nent it  can  be  freed  of  it  by  evolution  ;  if  its  exponent  be  a  positive 
fraction  it  can  be  freed  of  it  by  extracting  the  root  indicated  by  the 
numerator  of  the  exponent,  and  involving  this  root  to  the  power  indi- 
cated by  the  denominator.  If  the  exponent  of  the  unknown  quantity 
is  negative  it  can  be  rendered  positive  by  multiplying  the  equation  by 
it  with  a  numerically  equal  positive  exponent,     q.  e.  d. 

EXAMPLES. 

1.  Solve  y  — •  —  =  ttV. 

SuG.  2/3  =  8.  .  • .  y  =  2.  Why  not  put  the  +  sign  before 
the  2?  ~ 

/>  q 

2-  Solve  ^j^  =  -  gj^rr^.  BesuU,  y  =  -4. 

Query. — Can  the  root  of  a  Pure  Cubic  Equation  be  imaginary? 
Why? 

3.  Solve  3a;^  —  5  =  2A  Result,  x  =  125. 

4.  Solve  ^  +  1  =  5(^ 1).  Eesult,  x  =  32. 


—  2.      5       2  25 

5.  Solve  12^    3  _|_     =        _j_  Eesult,  x  =  27. 

o  -4  if 


358        EQUATIONS  OF  OTHER  DEGREES 

6.  Given    {x  +  {Ux  +  lla^)^^  =x^  +  a^  to   find  the 
value  of  X.  Result,  x  =  16a. 


7.  Solve  3aa;»  =  2aa7'»  +  56.  Result, 


-"Nl(?y- 


8.  Solve  a  —  l=6a;»       —  a;™     . 

a  — I 


Result, 


-(^ef)--"- 


9.  Solve  x^  =  27.     Also  x^  =  4.     Also  ?/^  =  32. 

Roots,  81,  +  32,  and  8. 

QuEKY. — Why  the  -f  sign  in  one  case  and  not  in  the  others  ? 


108,  J*rop.  2.  Any  equation  containing  one  unknoum 
quantity  affected  with  only  two  different  exponents,  one  of  which 
is  twice  the  other,  can  he  solved  as  an  Affected  Quadratic. 

Dem.  — Let  m  represent  any  number,  positive  or  negative,  integral  or 
fractional  ;  then  the  two  exponents  will  be  represented  by  m  and  2m  ; 
and  the  equation  can  be  reduced  to  the  form  x-""  +  px'^  =  q.  Now 
let  y  =  x^,  and  y"  =  x"^^,  whatever  m  may  be.     Substituting  we  have 

y2  ^  py  =  q^  whence  ?/=— |-  -+.   .|^+^-     ^^^   V  =  a;'"  ;    hence 

EXAMPLES. 

1.   Solve  Sx^  +  A2x^  =  3321. 

3. 

Solution. — Let  y  =  x^,  whence  ?/-  =  ic^  ;  and  Sy-  -f-  42y  =  3321. 
From  this  y  =  27,  and  —  41.  Taking  the  first  value,  x^  =  27.  .  • .  a 
=  9.  Taking  the  second  a;^  ==  —  41  .  • .  x  =  1^16817  We  therefore 
find  that  aj  =  9,  and  ^1681. 


SOLVED   AS   QUADRATICS.  359 

2.  Solve  x^  +  Ix^  =  44.         ^  =  +  8,  or  +   (—  11  )i 
QuEKY. — How  many  values  ?    "Which  are  imaginary  ? 

3.  Solve  Ax^  +  j;«  =  39.  x=  729,  and  (^)'* 

4.  Solve  3^6  _|_  42^3  =  3321.  x  =  d,  and  —  '^il. 

5.  Solve  —  +  2  =  ^  x=4.,  and  ^A^i 

ScH.  —It  is  not  necessary  to  substitute  another  letter  for  the  unknown 
quantity  as  given  in  such  examples.      Thus,   in  Ex.  3,  doubling,  to 

avoid  fractions,  Sx'^  -f  2x^  =  78.     Completing  the  square  8"x^   -f-  (m) 

+  1  =  8-78  +  1  =  625.     Extracting  root8x^  + 1=  +  25.  a:«  =  3, 

13  /13  \6 

and  —  ^.     .  • .  X  =  729,  and  (  — -  j  . 

[Note.  —Solve  the  next  six  without  substituting.] 

6.  Solve  x^o  _j_  six^  =32.  or  =  1,  and  —  2. 

1  1 

7.  Solve  a;"  +  13^'^  =  14.  ^  =  1^  and  (—  14)'". 


8.  Solve  3^2  _  4^1  ^  7. 


9.  Solve  3x  +  2v/^  =  1. 


/7\^ 


Suggestions.     (3)-x  -|-  (?n)  +  1=4.     .  • .  -/x  =  -^  and  —  1. 
X  =  ^,  and  1 . 

10.  Solve  ^2x  —  Ix  =  —  52.  .x  =  8,  and  ^%K 

11.  Solve  X  +  2v/a^  +  c  =  0. 

X  =  { —  s/a   -4-  v^a  —  c}2. 

jg  _  1  +  ^2 

12.  Solve  x^ X  —  — 7=— • 

yx  V  X 

X  =.±  v/l   -f  1^27 


360  EQUATIONS  OF  OTHER  DEGREES 


SuG. — The  first  member  may  be  written — .     Hence  drop- 

ping  the  denominator,  x/x,   and  squaring,  6ic"^  ■ —  x^  =  1  -\-  2x^  -{-  x*. 


FOR  REVIEW  OR  ADVAINCED  COURSE. 

100,  J^VOJ}*  3>  Equations  may  frequently  he  put  in  the 
form  of  a  quadratic  by  a  judicious  gi^ouping  of  terms  contain- 
ing the  unknoum  quantity,  so  that  one  group  shall  be  the  square 
root  of  the  other. 

Dem.  — This  proposition  will  be  established  by  a  few  examples,  as  it  is 
not  a  general  truth,  but  only  points  out  a  sjiecial  method. 

EXAMPLES. 

1.  Solve  2x'-  +  3^  —  5  \/2x'^-\-'6x -{- d  +  3  =  0. 

Solution.  — Add  6  to  each  member  and  arrange  thus,  (2x2  -j-  3x  -f-  9) — 

5 (2x2  +  3x -I- 9 ) ^  =  6.  Put  (2x-2  -{•3x-\-9f  =y,  and  the  equation  be- 
comes t/'^  ^  5^/ =  6.  Whence  t/  =  6,  and  — 1.  Taking  j/=*6,  2x--j- 
3x  -f  9  =  36.     Whence  x  =  3,  and  —  4i     Taking  y  =  —  1,  2x2  -f  3x  -|-  9 

=  1.     Whence  x  = = • 

4 

2.  Given  {2x  +  6)i  +  (2a;  +  6)4  =  6,  to  find  the  values  of  x. 

SuG.— Put  2/  =  (2x  +  6)  ;  whence  y^-  -{- y  =  6,  y  =  2  and  —  3. 
.-.  2x-f6  =  16,  andalso2x  +  6  =  81.     x  =  5,  and37i 

QuEBY.  —Will  the  value  x  =  37 ^  verify  ?     Why  ? 

Ans. — Since  (2x -J- 6)  and(2x-j-6)  are  even  roots,  their  signs  are 
strictly  ambiguous,  though  not  so  expressed  in  the  example.  Substi- 
tuting for  X,  372,  the  equation  becomes  (81)    +  (81)^  =6.     If  now  we 

regard  (81)  =  —  3,  as  it  is,  as  really  as  it  is  -|-3,  the  value  verifies. 
Such  cases  are  frequent. 

3.  Given  (x -j- 12)^  ==  6  —  {x -\- 12)^  to  find  the  values  of 
X,  and  verify  both  values. 


SOLVED   AS  QUADRATICS.                              361 

1  2 

4.  Given -—  =  i  +  — —-  to   find   the   values 

of  X.  Values,  x  =  3,  and  1. 

SUG.— Put  {2x  —  4)2  =  y. 


5.  Solve  x-\-5  —  vx  +5  =  6.  J7  =  4,  and  —  1. 


6.  Solve  2^x-'  —  'Sx+ll==x^  —  3x 


SuG.    xi  —  3x-{-8  —  2  Vx^  —  3a;  -f  11  =  0.     Add  3  to  each  member 
id  x2  —  3x  -f  11  —  2v/x^-  —  3x  -Pli  =3.  ,  Put  v 
!  — 2t/  =  3.     .-.  x  =  2,  1,  or  i(3  +  v/— 31.) 

7.  Solve  (x-^  _  9)2  =  3  +  ll(a;2  _  2). 


:c  =  4-  5,  and  +   2. 

8.  Solve  (x  +  -V  +  X  =  42  —  -. 

\         ^/  X 

07  =  4,  2,  and  ^(—  7  +  ^/vf). 

/  1  \2 

9.  Solve  x4l  +  _  j  _  (3;r2  +  a;)  =  70. 

SuG.    x'(l  +  —)~  =  i(3x2  -j-a;)2.  Hence  the  equation  may  be  written 
(3x2  _|_  ic)2  —  9(3xi  +  .X)  =  630. 


Besults,  X  =  S,  —  8-^,  and  ^(—  1  +  ^—  251). 

10.  Solve —  =  —— — .  Boots,  x  =  4,  and  1, 

X — v^j?  ^ 

/  —  1  X  —  V  X 

SuG. — Divide    by  x  +  vx,    and = — .     Hence   4  = 

X  —  \/x  ^ 

(x  —  >/x)2,  or  X  —  \/x  =  -f  2. 

110  •  CoR. — The  foem  of  the  compound  term  may  some- 
times be  found  by  transposing  all  the  terms  to  the  first  member, 
arranging  them  with  reference  to  the  unknown  quantity,  and 
extracting  the  square  root.  In  trying  this  expedient,  if  the 
highest  exponent  is  not  even  it  must  be  made  so  by  multiplying 
the  equation  by  the  unknown  quantity.  In  like  manner  the 
coefficient  of  this  term  is  to  be  made  a  perfect  square.      When 


3B2         EQUATIONS  OF  OTHER  DEGREES 

the  process  of  extracting  the  root  terminates,  if  the  root  found 
can  be  detected  as  a  part,  or  factor,  or  factor  of  a  part  of  the 
'emainder,  the  root  may  be  the  polynomial  term. 

.-,    c  .       ^         30       12  +  ^^         7         ,, 

11.  Solve---  +  -^^=--  +  H. 

Solution. — Cleared  of  fractions,  transposed  and  arranged,  this  be- 
comes Sx-i  — 42x^5 -}-■  168x2  — 147x  — 180  =  0.     Dividing  by3,  x^  — 14x3 
-f  56x2  —  49x  —  60  =  0. 
Extracting  square  root  x"  —  14x^  -j-  56x2  —  49x  —  60 1x2  —  7x. 

x" 
2x2  _  7a.|  _  i4a;3  _^  ^q^^ 

—  14x3  -{-  49x2 

7x2"-- 49x  _  60 
7(x2  — 7x)— 60 
After  obtaining  two  terms  of  the  root  we  observe  the  root  itself  as  a  fac- 
tor in  part  of  the  remainder.     The  polynomial  is  therefore  (x2  —  7x)2 
-f-  7(x2  —  7x)  —  60  =  0.     This    is    solved  as  before,     x  =  4,  3,  and 
^(7  +  Vm). 

12.  Solve  x^  —  12x^  +  4Ax^  —  48^  =  9009. 

One  value  of  x  is  IS. 

13.  Solve  J73  —  6j:2  4-  lla:  _  6  =  0.  x  =  1,  2,  and  3. 
14  Solve   4:X*  +  ^  =  4^3  ^  33. 


x  =  2,—^,  andi(l  +  \/  — 43)- 


15.  Solve  X*  —  2x3  +  ^  =  a.  ^  ^  |.  _|_  V|  ^  v/a  +  i 

16.  Solve  X*  -\-  x^  —  4:X^  -{-  X  -{-  1  =  0. 

J7  =  1,  and = . 

Solution. — Dividing  by  x2,  x2  -f-  x  —  4  -j }-  -  =  0,  which  may  be 

written  x2  +  2  -f  -  -f  x  -f  -  =  6,  or  (x  -f  -)  -|-  f'^^  +  - )  =  f>-    Whence 
the  compound  term  appears. 


SOLVED  AS   QUADRATICS.  363 

[Note. — The  discovery  of  such  devices  as  are  required  for  the  solu- 
tion of  this  equation,  is  quite  beyond  the  skill  of  any  but  expert  alge- 
braists. ] 


111,  JPvop,  4.   When  an  equation  is  reduced  to  the  form 

x"  +  Ax«-'  +  Bx"-'  +  Cx"-^ +L  =  0,  the  roots,  with  their 

signs  changed,  are  factors  of  the  absolute  (Jcnown)  term,  L. 

Dem. — 1st.  The  equation  being  in  this  form,  if  a  is  a  root,  the  equa- 
tion is  divisible  by  x  —  a.  For,  suppose  upon  trial  ic  —  a  goes  into  the 
polynomial  x"  \-  A.t"  — ^  -(-,  etc.,  Q  times  voiih  a  remainder  E,.  (Q  rep- 
resents any  series  of  terms  which  may  arise  from  such  a  division,  and 
K,  any  remainder. )  Now,  since  the  quotient  multipHed  by  the  divisor, 
-}-  the  remainder  equals  the  dividend,  we  have  (x  —  a)  Q  -}-  11=  x"  4" 

Aic"~^  -\-  Bx"-2  -f-  Cx"— ^ \-L.    But  this  polynomial  =  0.    Hence 

(a;  —  a)Q  -f-  E  =  0.  Now,  by  hypothesis  a  is  a  root,  and  consequently 
X  —  a  =  0,     Whence  E,  =  0,  or  there  is  no  remainder. 

2nd.  If  now  x  — a  exactly  divides  x"  -|- Ax"  -  ^  -|-^^"  ~  ^  +  Ca;"  ~  ^ |-L, 

a  must  exactly  divide  L,  as  readily  appears  from  considering  the  pro- 
cess of  division.  Hence  —a  is  a  factor  of  L,  a  being  a  root  of  the 
equation.     Q.  E.  D. 

[Note. — In  most  of  our  better  text  books  there  will  be  found  special 
methods  for  reducing  some  forms  of  higher  equations  (usually  cubics, 
it  will  be  observed, )  to  quadratics,  by  separating  and  arranging  the 
terms,  adding  some  term,  multiplying  or  dividing  by  some  number, 
BO  as  to  make  some  binomial  factor  appear  which  can  be  divided  out, 
or  so  that  both  members  shall  become  perfect  squares.  This  method 
is  always  difficult  of  application,  depending,  as  it  does,  solely  upon 
the  ingenuity  of  the  student.  The  binomial  factor  sought  by  these  vieans 
is  readily  found  by  the  principle  above  given.  The  difficulty  in  always 
discovering  a  root,  on  this  principle,  arises  from  the  fact  that  the  num- 
ber of  factors  of  L,  including  fractional  and  imaginary  ones,  is  infinite. 
So  that  this  method  will  be  of  service  only  when  some  one  or  more  of 
the  roots  is  an  easily  recognized  factor  (usually  an  integral  one)  of  L. 
But,  in  this  respect,  the  method  is  equally  as  good  as  the  cumbrous  ones 
usually  given,  since,  in  cases  where  we  could  not  recognize  the  root  in 
this  way,  it  would  be  little  less  than  hopeless  to  attempt  to  do  it  in  the 
other.  In  fact,  the  examples  ordinarily  given  to  illustrate  the  expedi- 
ents referred  to,  are  here  given,  and  solved  by  a  simple  and  uniform 
method.] 


364  EQUATIONS   OF   OTHER  DEGREES 

EXAMPLES. 

1.  Solve  1x^  —  68a:  =  32  —  IT^^. 

17x3 

Solution. — This  may  be   written   x^  -| 34a;  —  16  =  0.    By 

2 

Pkop.  4  the  roots  of  this  equation  are  factors  of  16.  We  therefore  try- 
in  order  ±\  ±%  ±^,  ±'^,  ±  16,  (all  the  integral  factors  of  16)  till 
we  find  whether  there  is  an  integral  root.  We  see  at  once  that  neither 
-f- 1  nor  —  1  satisfies  the  equation.     Trying  -f-  2  we  find  it  is  a  root. 

17a;3 
Hence  x  —  2  is  the  divisor  sought.     Dividing  x^  -J — 34a;  — 16=0, 

21 
by  X  —  2,  we  have  x^  -f  —  x^  -f-  21x  +  8  =  0.     But  as  —  2  satisfies  the 

a 

21 
equation  as  well  as  -j-  2,  x  +  2  is  a  divisor.     Dividing  x^  -f-  —  x^  -f-  21x 

17 

4-8  =  0byx-j-2we  have  x'^  +  —  x  +  4  =  0.       From   this   equation 

a 

x  =  —  8,  and  —  h-     The  roots  therefore  are  2,  —  2,  —  8,  —  i. 

2.  Solve  ;r—l=  2+-?=. 

\^x 
SuG. — Put  y  =  \/x,  and  clearing  of  fractions  and  transposing, 
y3  —  3y  —  2  =  0.  If  there  are  integral  roots  of  this  equation  they  are 
4-  1,  or  4-  2.  -f-1  cloes  not  satisfy  the  equation  and  hence  is  not  a 
root.  But  —  1  does.  Hence  y  -j-1  is  a.  divisor.  Dividing,  we  get 
t/2  —  y  —  2  =  0.  Whence  j/  =  2,  or — 1.  ".  •.  The  roots  of  the  equa- 
tion y^  —  ^y  —  2^0  are  — 1,2,  and  —  1,  there  being  two  equal  roots. 
Now  as  2/  =  \/x,  X  =  1,  and  4. 

3.  Solve  x^  —  3x^-}-x  +  2  =  0. 

The  irrational  roots  are  ^{1  -f  n^5). 

4.  Solve  x^  =  Qx  -\-  9. 


Imaginary  roots,  ^{  —  3  -f-  y  —  3). 

[Note. — The  simpler  roots  are  not  given  in  these  results,  since  they 
would  at  once  indicate  the  factor  to  be  divided  out  of  the  equation.  ] 

5.  Solve  2x^  —  x^  =  1,  or  x^  —  ^x'^  —  i  =  0.     (  +  1  is  a 
factor  of  — -1-).         Imaginary  roots,  ^( — 1  -f  v^ — 7). 

2 

6.  Solve  x^  —  3-  =  H'y  or  x^  —  -\^-x  —  f  -=  0.     ( +  1  and 

ox 

4-  f  are  factors  of  the  absolute  term.) 

Surd  roots,  ^{1  ±  v^). 


365 


SOLVED   AS   QUADRATICS. 

7.  Solve  X  -{-Ix^  ==  22.      (Put  x'^  =  y.)  

Imaginary  Boots,  29  +  1^ — 10. 

8.  Solve  x^  +  -i/a;3  —  39a;  =  81. 

Imaginary  Boots,  ^{ — 13  -f  V  — 155). 

^    ^  .  12  +  8x/x 

9.  Solve  X  = — . 

X  —  5 

Imaginary  Boots,  ^( — 3  +  v — 7). 

Suggestion. — See  Ex.  2  above, 

,^    ^  ,      49^2        48        ,^        ^6 
10.  Solve  — 49  =  9  +  -. 

4:  X'^  X 


Surd  Boots,  }{—  3  +  v^93). 


232  24:  192 

Sua — This  put  in  the  required  form  is  x* —  x^  — —a;  +    377  =  0' 

Examine  for  integral  roots  by  trying  4-  1,  and  +  2,  etc.  The /or?n  of 
the  absolute  term  suggests  also  a  fractional  root.  To  test  for  this  we 
would  try  fractions  with  7  for  denominator.  These  processes  may  be 
a  Httle  tedious,  but  there  is  method  in  them.     But  suppose  this  ques- 

232  24  192 

tion  given  in  the  form  x^  —  Iq""*'^'  =  To"'^  —  Tq~'     -^^  what  possible 

means  could  the  pupil  discover  that  it  must  be  put  in  the  form  given  in 

the  example,  and  then  the  square  of  both  members  completed  by 

1                                                                               49x-                    49 
adding  — ^  to  each  member,  and  writing  the  result,  — ^ 49  -^ 

XX- 

\.     c  -,      nrr  841         17         232  1     ,    r 

11.  Solve27.2---+-=— -— +5. 

2  232  840 

SuG. — From  the  equation  we  have  x^  -{-ZTr""^^ 31"^ oi~  =^- 

ol  ol  ol  / 

Searching  for  integral  roots  we  readily  find  2,  and  dividing  out  the 

326  420 

factor  x  —  2,  have  cc^  -f-  2x-i  -| — -x   +  --—  =  0.     To  find  a  root  of 

81  ol 

14  14 

this  by  inspection  would  be  a  hopeless  task.     It  is — ,  and  x  -| — — 

is  the  divisor  which  reduces  the  equation  to  a  quadratic  whose  roots 

are  9  (—  2  +  v'-  266> 


366  EQUATIONS   SOLVED   AS    QUADRATICS. 

The  method  of  solving  this  equation  by  completing  the  square  of 
both  members  is,  to  multiply  by  3,  transpose  — j-  and  —  (not  uniting 
them),  and  add  one  to  each  member,   thus  deducing  81ic2  -^  18  -f- 

1  841  232 

~    =  — ~     -j-    — —   -f-  16.      Extracting  the   square  root  of  both 

members  9x-\ —  =  ±  ( ^  ^r 

,^    ^  ,      18       81  —  X-'        ^2_65 

12.  Solve h  — t: •  =  7r\ ' 

SuG. — Putting  this  in  the  form  required  the  absolute  term  is 
—  1296  ;  the  integral  factors  of  which  are  very  numerous.  But  trying 
±  1'  ±  2,  ±  3,  4-  4,  +  6,  4-  8,  +  9,  we  find  —  4  and  9  to  be  two 
of  the  roots. 

The  method  of  solving  this  by  putting  it  in  form  so  as  to  complete 

8t         81 
the  square  of  both  members,  is  to  multiply  by  2,  and  add  -—  -f-  ^^ 

36        So 

36        18       9         x^        2x        4 
to  both  members,  obtaining  ^  -| -\-  —  =  .^  -\-  — — f-  — .         Ex- 

6        3  /  X       2  \ 

tracting  root  --j-=4-(-4--).      Whence  x  =  9,  —  4,  —  4,  and 

—  9,    there  being  two  roots  —  4. 

3  _^  4v/^ 

13.  Given  x  —  3  = to  find  the  roots. 


SuG. — The  roots  being  all  surds  and  imaginaries  in  this  equation 
cannot  be  found  by  the  principle  in  the  proposition.  The  following 
special  expedient  may  be  resorted  to  : 

Clear  of  fractions  and  add  x  -f- 1  to  each  member  and  xfi  —  1x  -\-\ 
=  4  -j-  4v/x  +  X. 

14.  Solve  the  following  by  the  principle  in  the  proposition: 


jr2  —  1        3 


x  —  \    _  3       15   J7  +  1 
^  +  5  X       X    X  +  5' 

^4  4.  ^3  _  4^2  _4_  ^  ^  1  =  0.     (See  Ex.  16,  Prop.  3.) 

[Note. — Many  of  the  examples  unde*-  Prop.  3  can  be  readily  solved 
in  this  manner.  ] 


SIMULTANEOUS  EQUATIONS.  367 


SECTION  IV. 

Sunultaneous  Equations  of  the  Second  Degree  between 
two  Unknown  Quantities. 

112,  I^VOjy.  1,  Two  equations^  between  two  unknown 
quantities,  one  of  the  second  degree  and  the  other  of  the  first, 
may  always  be  solved  as  a  quadratic. 

Dem. — The  general  form  of  a  Quadratic  Equation  between  two  un- 
known quantities  is 

ax^  4-  hxy  +  cy^  +  cZx  +  ey  +  /  =  0, 
since  in  every  such  equation  all  the  terms  in  a;-  can  be  collected  into 
one,  and  its  coefficient  represented  by  a  ;  all  those  in  xy  can  also  be 
collected  into  one,  and  its  coefficient  represented  by  b,  etc .,  etc. 

The  general  form  of  an  equation  of  the  First  Degree  between  two 
variables  is 

a'x  -f-  h'y  -\-  c'  =  0. 

J)'lJ  (;' 

Now,  from  the  latter  x  = ,  which  substituted  in  the  for- 

a 

mer  gives  no  term  containing  a  higher  power  of  y  than  the  second, 

and  hence  the  resulting  equation  is  a  quadratic,     q.  e.  d. 

EXAMPLES. 

X  —  y  ^  +  3i/ 

1.  Griven  X  ■ — ~  =  4,  and  y  —  r  =  1- 

SuG. — From  the  first  a;  =  8  —  y.      Substituting  this  value  of  x  in 
8  —  V  +  3y  8  4-  2v 

the  second,  we  have  y  —  —  — = — ■; — -  =  1,  or  v  —  tt. =^  1 ;  whence 

^       8—  v-f-2  ^       KJ  —  2/ 

yz  —  9y  =  —  18,  and  y  =  6  and  3. 

2.  Given  x  +  y  =  1,  and  x^  +  2?/2  =  34.     Verify. 

3.  Given  — | —  =  2,  and  x  -}-  y  =  2.     Verify. 

X       y 

4.  Given  x  -\-  y  =  100,  and  xy  =  2400. 

1         1         14- 

5.  Given  2^  +  3v  =  37,  and  -  +  -  =  — . 

X        y         4:5 

6.  Given  2x^  -r  xy  —  5y^  =  20,  and  2x  —  Sy  =  1. 


368  SIMULTANEOUS   QUADRATIC   EQUATIONS 

7.  Given  x  -^\  =  — ^-^,  and  —-L-'l  ==,  —J, 

-^  3  X  2 

8.  Given  .ly  +  .125x  =  y  —  x,  and  ?/  —  .5x  ^=  .  15xy  — ■ 
Sx.  Besults,  X  ^  0,  and  4;  and  ^  =  0,  and  5. 

9.  Given  .3^  +  .125y  =  3x  —  y,  and  dx  —  .5y  =  2.25xy 
+  3i/. 

Results,  a;  =  0,  and  —  1;  and  3/  =  0,  and  —  2f. 


113  •  ^Top*  2,  In  general,  the  solution  of  two  quadrat- 
ics between  two  unknown  quantities,  requires  the  solution  of  a 
biquadratic. 

Dem. — Two  General  Equations  between  two  unknown  quantities 
have  the  forms 

(1)  ax2  4-  hxy  -f-  ct/2  _|_  dx  +  ey  +  /  =  0,  and 

(2)  a'a;2  _|_  h'xy  +  cY  +  d'x  +  e'y  +/  =  0. 

From  (1),  .=_^-^  +^  l(!i^)l_  2^1  +  ^05. 

Now,  to  substitute  this  value  of  cc  in  equation  (2),  it  must  be  squared, 
and  also,  in  another  term,  multiplied  by  y,  either  of  which  operations 
produce  rational  terms  containing  j/^,  and  a  radical  of  the  second  degree. 
Then,  to  free  the  resulting  equation  of  radicals  will  require  the 
squaring  of  terms  containing  y^,  which  will  give  terms  in  y*,  as  well 
as  other  terms.     Q.  e.  d, 

[Note. — Since  it  is  not  the  purpose  of  this  treatise  to  embrace  the 
resolution  of  the  higher  equations,  only  such  special  cases  of  Simulta- 
neous Quadratics  with  two  unknown  quantities,  will  be  introduced,  as 
can  be  resolved  by  the  methods  of  quadratics .] 


114.  I*voj)»  3,  Two  Homogeneous  Quadratic  Equations 
between  two  unknown  quantities  can  always  be  solved  by  the 
method  of  quadratics,  by  substituting  far  one  of  the  unknown 
quantities  the  product  of  a  new  unknown  quantity  into  the 
other. 

Dee. — A  Homogeneous  Equation  is  one  in  which  each  term  contain? 
the  same  number  of  factors  of  the  unknown  quantities.  2x2  —  3^;^  — 
j/2  =  16  is  homogeneous,     dx-  —  2?/  -\-  y-  =  10  is  not  homogeneous. 


WITH   TWO   UNKNOWN   QUANTITIES.  369 

Dem.  — The  truth  of  this  proposition  will  be  more  readily  appre- 
hended by  means  of  a  particular  example.  Taking  the  two  homoge- 
neous equations  x'^  —  ^V  -\-  V'  =  21,  and  y"^  —  ^xy  -|-  15  =  0.  Let 
X  =1  vy,  V  being  a  new  unknown  quantity,  called  an  auxiliary,  whose 
value  is  to  be  determined.  Substituting  in  the  given  equations,  we 
have  v^y-  —  vy^  -}-  t/2  =  21,  and  y^  —  2y?/'  =  -^  15.     From  these  we 

21  15 

find  y-  = -— ,    and    y-   =    r , .     Equating   these    values 

^         v^  —  v-\-l  ^  2u  —  1         ^  ^ 

21  15 

of  y2,  — — ,-—  =  -; ;  whence  42t;  —  21  =  15u2  _  ISu  +  15. 

u^  —  V  -j- 1       2i;  —  1 

This  latter  equation  is  an  aflfected  quadratic,  which  solved  for  v,  gives 

V  =  3,  and  |,     Knowing  the  values  of  v  we  readily  determine  those  of 

15  — 

y  from  y-  = -,  and  find  y  =  4.  v^3  when  v  =  3,  and  y  =    +  5 

when  u  =  f .  Finally  as  x  =  vy,  its  values  are  x  =  4-  3v/3,  and  4-  4. 
By  observing  the  substitution  of  vy  for  x  in  this  solution  it  is  seen 
that  it  brings  the  square  of  y  in  every  term  containing  the  unlcnown 
quantities,  in  each  equation,  and  hence  enables  us  to  find  two  values 
of  t/2  in  terms  of  v.  It  is  easy  to  see  that  this  will  be  the  case  in  any 
homogeneous  quadratic  with  two  unknown  quantities,  for  we  have  in 
fact,  in  the  first  of  the  given  equations,  all  the  variety  of  terms  which 
such  an  equation  can  contain.  Again,  that  the  equation  in  v  vnll  not 
be  higher  than  the  second  degree  is  evident,  since  the  values  of  2/2  con- 
sist of  known  quantities  for  numerators,  and  can  have  denominators 
of  only  the  second,  or  second  and  first  degrees  with  reference  to  v. 
Whence  v  can  always  be  determined  by  the  method  of  quadratics  ;  and 
being  determined,  the  value  of  y  is  obtained  from  a  pure  quadratic 

{y-  = -,  in  this  case),  and  that  of  x  from  a  simple  equation  (» 

=  vy  in  this  case). 

EXAMPLES. 

1.  Given  Sx-  +  ^y  =  18,  and  4?/^  +  dxy  =  54. 

Boots,  X  =  ^  2,  and  +  2  v^S  ;  and  y  =  _}_   3,  and 
+  3v/3. 

2.  Given  x-^  -\-  xy  +  2?/2  =  74,  and  2x^  -f  2xy  +  y'  =  73- 

Hoots,  ^  =  -j-  3,  and  +  8  ;  and  y  =  +  5. 

3.  Given  x'^  +  Sxy  =  54,  and  xy  +  4y2  =  115. 

Boots,  j7  =  4-  3,  and  +  36 ;  and2/=  -f  5,  and  -f  11-^. 


370  SIMULTANEOUS   QUADRATIC   EQUATIONS 

4.  Given  2x^  +  Sxy  =  26,  and  dy^  +  ^^y  =  39. 
Boots,  X  ^  -^  2,  and  y  =  ±  ^'     The  other  roots  areoo- 

6.  Given  x^  —  4?/2  =  9,  and  xy  +  2^/2  =  3. 

j?oo/s,  ^  ==  i  --== ;  and  y  =  ±  -y==.     The  other 

roots  are  oo- 

6.  Given  Sx^^-i-xy  —  9  =  9,  and  4?/2  +  3xy  —  4  =  50. 
(See  Ex  1.)  Boots,  x=  ±2,  and  y  =  +  3. 

7.  Given  x"^  —  ^y  =  70,  and  xy  —  y~  =  12. 

8.  Given  x^  -{-  xy  =  84,  and  x'^  —  y-  =  24. 


FOR  REVIEW  OR  ADVANCED  COURSE. 

US*  J^TOp,  4.  When  the  unknown  quantities  are  simi- 
larly involved  in  two  quadratic,  or  even  higher  equations,  the 
solution  can  often  he  effected  as  a  quadratic,  by  substituting  for 
one  of  the  unknown  quantities  the  sum  of  two  others,  and  for 
the  other  unknown  quantity  the  difference  of  these  new  quan- 
tities. 

[Note. — As  this  and  the  following  are  merely  special  expedients,  they 
need  no  demonstrations  other  than  is  furnished  by  applying  them  to 
examples.  ] 

EXAMPLES. 

1.  Given  x^  -{-  y^  =  52,  and  x  -}-  y  -{-  xy  =  M. 
Solution.  — In  these  equations  x  and  y  are  similarly  involved,  and 

hence  I  try  the  expedient  of  putting  a;  =  m  -)-  n,  and  y  =  m  —  n,  whence 
X-  4-  y^  =  2m2  -f-  2?z'-  =  52,  and  x  -\-  y  -\-  xy  =  2m  -{-  m^  —  n^  =  34.  Now, 
from  the  two  equations  m2  -f-  n-  =  26,    and 

2m  -f-  wi^  —  n^  =  34,    by  adding 
I  have  2m  -f-  2m2    =      60,    whence  I  find  m  =  5,  and 

—  6.  Substituting  these  values  in  m^  -j-  n'^  =  26,  n  =  -i-  1,  and 
4-  s/ —  10.  Whence  the  real  values  of  x  are  found  to  be  6,  and  4  ;  and 
of  y,  4,  and  6. 

2.  Given  x^  -\-  x  -{-  y  ^=  18  —  y^,  and  xy  =  6. 

Bational  roots,  x  =  3,  and  2;  and  y  =  2,  and  3. 


WITH   TWO   UNKNOWN   QUANTITIES*  371 

3.  Given  —^•-  + ^  =  — ,  and  x'^  +  y^  =  45. 

X  —  y      X  -{-  y         o 

SuG. — Using  the  same  notation  as  above, 1 =  -rrr  ^^^  ^m'^  4- 

^  n        m        3 

2n2  =  45  ;    whence  3m-  -j-  3n-  =  lOmw,    and   we  have  imn  =  27,   or 

27  9  3  3  9 

m  =  — .     m=  +  -,  and    +  -  ;  7i  =  +  -,   and  +  ..      The   roots   are 

4n  ~2  ~"2  ~2  ~  ^ 

x=±  6;  and  2/  =  +  3. 

4  Given  4(^  -|-  y)  =  3a;t/,  and  x -\- y  +  x^ -]- y"- =  26. 

Boots,  x  =  4,,  and  2  ;  2/  ==  2,  and  4.  Also  x  =  —  i^- 
+  ^v/377,  and  7/  =  —  -y.  =F  iv/377: 
Def.  and  Sch. — It  will  be  observed  that  the  above  equations  are  of 
the  second  degree,  and  that  they  have  the  unknown  quantities  simi-  ^ 
larly  involved  ;  that  is,  in  the  last,  for  example,  in  the  first  member 
there  is  -f-  ^^,  and  also  -f-  %  ;  in  the  second  member  x  and  y  are  mul- 
tiplied together  ;  in  the  second  equation  there  is  -j-  x,  also  -|-  y  ;  there 
is  -f-  x^,  and  also  -f-  2/^-    Such  equations  can  usually  be  readily  solved  in 

5  1 

this  manner.     But  the  equations  x-  -\-  y-  =  -xy,  and  x  —  y  =  -xy  have 

the  unknown  quantities  similarly  involved  in  the  first  but  dissimilarly 
in  the  second.  There  is  -f-  ^  in  the  second,  but  no  -|-  y,  hence  they 
are  not  similarly  involved.  Whether  the  solution  of  such  equations 
will  be  facilitated  by  this  expedient  can  only  be  ascertained  by  trial. 
In  this  case  the  expedient  will  be  found  successful. 

5.  Given  x'^  -]-  y^  =  ^xy,  and  x  —  y  =  \xy. 

Boots,  or  =  0,  4,  and  —  2 ;  y  =  0,  2,  and  —  4. 

6.  Given  x^  +  xy-i-  Ay'^  =  6,  and  S^-^  +  8if  =  14. 

Boots,  07=  4-  2,  and  +  |yiO  ;  and  ?/  =  +  ^,   and 

±  1^10. 

7.  Given  x-^  —  4?/2  =  9,  and  xy  +  2rj^  ^d. 

SuG. — The  student  will  find  by  experiment  that  the  above  expedient 
is  of  no  service  in  this  example.     The  example  is  readily  solved  by 
finding  the  value  of  x  from  the  first  equation  and  substituting  it  in  the  . 
second,  thus  obtaining  ys/Ay^-^-d  =3  —  2f,  or  iy^  +  dy^  =d  —  12y^ 

-\-iy*.  Whence  21y*  =  9,  and  y  =  +  I--  ^^  ^^^  ecpuations  can  be 
treated  as  in  [^114:),  they  being  homogeneous. 

8.  Given  -  -\- K'  =  18,  and  j;  +  y  ==  12. 

V         X 


372  SIMULTANEOUS    QUADRATIC    EQUATIONS 

SuG. — In  these  equations  the  unknown  quantities  are  similarly  in- 
volved, and  although  the  first  is  of  the  third  degree  the  expedient  of 
the  proposition  is  successful.     x  =  8,  and  4  ;  and  y  =  4:,  and  8. 

ScH. — In  all  symmetrical  equations  the  value  of  the  unknovm  quan- 
tities must,  of  course,  be  the  same  numerically,  but  taken  in  the  re- 
verse order,  since  the  letters  can  change  places  in  the  equation  without 
altering  the  equations.  When,  therefore,  in  such  equations  the  values 
of  one  of  the  unknown  quantities  are  found  the  values  of  the  other 
are  known. 

9.  Given  x^  —  x-y^  -\-  y^  =  ±2,  and  x  —  xy  -{-  y  =  L 

SuG. — Putting  x  =  m-^n,  and  y^=m  —  n,  x^  -\-y^  =  1m^  -\-  2n^, 
and  x^y^  ={m  -f-  n)^{m  —  n)^  =  {m  -}-  n){m  —  n)  X  {m  -\-  n){m  —  n)  - 
=  (rn'^  _  n^)2.      Hence  2m^  -|-  2n2  —  (m^  —n^f  =19.     From  the 
second  equation,  n^  z=  4  -j-  wi^  —  2m. 

Boots,  X  =  -1  (9  +  \/73)  ;  and  ?/  =  ^(9  +  ^73). 

10.  Given  x  +  y  =  11,  and  x^  +  y^  =  407. 

Boots,  X  =  1,  and  4  ;  and  ?/  =  4,  and  7. 

11.  Given  x  —  ?/  =  3,  and  x"^  -{-  y^  =  641. 

Boots,  X  =  5,  and  —  2  ;  y  =  2,  and  —  5. 


SPECIAL  EXPEDIENTS. 

no*  Many  equations  of  other  degrees  than  the  second, 
and  which  do  not  fall  under  the  preceding  cases,  may  still 
be  solved  as  quadratics  by  means  of  special  artifices.  For 
these  artifices  the  student  must  depend  upon  his  own  inge- 
nuity, after  having  studied  some  examples  as  specimens. 
These  methods  are  so  restricted  and  special  that  it  is  not 
expedient  to  classify  them  ;  in  fact,  every  expert  algebraist 
is  constantly  developing  new  ones. 


1.  Given  x^  -{-  y'^  =^  5,  and  x^  -\-  y^  =  13. 


SuG. — In  case  of  fractional  exponents,  it  will  usually  be  found  expe- 
dient for  the  learner  to  put  the  unknown  quantities  with  the  lowest 
exponents,  equal  to  the  first  powers  of  new  unknown  quantities,  and 

thus  make  the  exponents  integral.    Thus,  putting  x*  =  7n,  and  y*  =  n, 

1  2, 

we  have  x^  =  m^,   and  ?/'  =   n'.      Whence  the  equations  becom(, 


T^^TH   TWO   UNKNOWN   QUANTITIES.  373 

tn  -^  n  =  5,  and  m-  -j-  n-  =  13.     These  equations  are  readily  solved 
by  methods  already  learned,  and  we  find  m  =  3,  and  2,  and  n  =  2, 

and  3.     Hence  x"*  =  3,  gives  a;  «=  81  ;  and  x*  =2,  gives  x  =16,  Also 

1/=*  =  2,  gives  ?/  =  8  ;  and  ^■'  =  3,  gives  y  =  27. 

2.  Given  ^^  +  y ^  =  6^  a^d  j;*  +  y^  =  126, 

i^oo^s,  it;  =  625,  and  1,  and  y  =  1,  and  3125. 

3.  Given  x^y^  =  2y\  and  8^^  —  y'^  =  14. 

4.  Given  x  —  y  =  v/^  4-  v/j^^  and  x^  —  2/^  =  37. 

SuG. — Observe  that  both  members  of  the  first  are  divisible  by  -s/x 
+  Vy,  giving  V'x  —  Vy  =  I.     x  =  16,  and  9  ;  i/=  9,  and  16. 

5.  Given  x^  -\-  x  -{-  y  ^  IS  —  t/2,  and  xy  =  6. 

SuG. — From  the  first,  by  adding  twice  the  second,  we  may  write 
x2+2xy4-y2  4_x-fy  =  30,  or(x  +  2/)*  +  (ic4-2/)  =  30.  .-.  x  + 
y  =  5,  and  —  6. 

6.  Given  x^  -\-  y^  ^  52,  and  x  -\-  y  -\-  xy  =  34. 

Boots,  X  =  6,  4:,  and  —  6  -f  y/ — 10  ;  ?/  =  4,  6,  and 
—  6  +  v/— 10- 

7.  Given  x^  -{■  y^  +  4=  >/x^  +  y2  =  45,  and  x^  _|_  ^4=  337. 

SuG, — In  the  first,  put  -^x*  -(-  y*  =  u,  whence  v*  -f-  4i;  =  45  ;  and 
y  ^=:  5,  and  —  9.     x  =  3,  and  4  ;  y  =  4,  and  3, 

X*  x^ 

8.  Given f-  2—  =  9|S   and  j;^  +  v'  =  65. 

X*  39 

SuG. — In  the  first,  put =  v,  whence  v^  -f-  2u  =  9  — , 

y  49 

Real  and  rational  roots,  x  ==  4,  y  =  7. 

9.  Given  x''  +  2xy  -\-  y^-  -^  2x  =  120  —  2y,  and  xy  —  if 
=  8. 

i^oo^s,  y  =  1,  4,  —  3  —  v/5  a^^l  —  3  +   v/5'; 
x=  9,  6,  —  9  +  v/5,  and  —  9  —  v^ 5. 


374  SIMULTANEOUS   EQUATIONS. 

10.  Given  ^72^2  ^=  I8O  —  8xy,  and  x  -\-  dy  =  11. 

Boots,  X  =  5,  and  6  ;  y  =  2,  and  f . 

11.  Given  x-^ -\- 3x -{- y  =  7^  —  2xy,  and  y^  -]-  3y -{- x  =  44 

SuG. — Add  the  two  equations  together,  and  proceed  as  before, 
a;  =  4,  16,  and  -  13  +^58"  ;  and  ?/  =  5,  —  7,  and  —  1  +  n/58. 

12.  Given  xy  +  xy^  =  12,  and  x  +  ^y^  =18. 

12  18 

SuG.— From  the  first,  x  =  — - — ; ;  and  from  the  Becond,  x  =  r— — . 

12                 18 
.'.    = .     Dividing  denominators  by  1  -4-  V»  and  nu- 

ya-hy)      1  +  2/'  y     -ry^ 

2  3 

merators  by  6,  we  have  -  = ;  whence  2  —  2v  4-  2v2  = 

y      1  —  .V  +  y-  y  ^    y 

3y.     Hence  x  =  2,  and  16  ;  and  y  ==2,  and  ^. 

13.  Given  x"^  -}-  xy  -\-  y^  =  26,  and  x^  +  x^j^  -\-  y^  =  364. 

SuG. — From  the  first,  x^  -\-y^  =2G  —  xy  ;  and  from  the  second,  by 
adding  x'-y-  to  both  members  and  extracting  the  square  root,  x^  -f-  y^ 
=  -v/y64:  +  x'^-  Equating  these  values  of  x'^  -\-  y^,  and  squaring, 
we  have  676  —  52xy  -j-  ^^y'^  =  364  -j- x'-y',  whence  xy  =  6.  Squaring 
this  and  adding  it  to  the  second,  and  extracting  the  square  root, 
x^  -\-  y'^  =  20.  .Also  subtracting  Sx'^y^  =  108  from  the  second,  and 
extracting  the  square  root,  x-  —  y'^  =  16.  Whence  x^  =  18,  and  y^ 
=  2. 

Another  Solution. — Dividing  the  second  by  the  first,  we  have  x* 
—  xy  -\-  y'^  =  14.  Subtracting  this  result  from  the  first,  we  have  2xy 
=  12.     Whence  the  solution  proceeds  as  above. 

[Note. — Though  this  field  is  illimitable,  it  is  not  thought  necessary 
for  the  learner  to  pursue  special  methods  farther,  inasmuch  as  what  is 
given  will  enable  him  to  catch  the  spirit  of  such  solutions,  and  no 
writer  in  discussing  a  problem  involving  processes  even  as  complex  as 
Bome  given  above,  would  fail  to  give  hints  at  his  methods  of  solution.] 


APPLICATIONS. 

[Note. — One  of  the  most  important  things  to  be  learned  from  the 
following  examples  is  several  devices  frequently  found  serviceable  in 
stating  a  problem,  wfhich  make  the  equations  arising  more  simple  and 
easy  of  solution.  These  devices  are  of  special  necessity  in  examples 
involving  progressions.] 


OF    THE    SECOND    DEGREE.  375 

1.  What  number  is  that  which  being  divided  by  the  pro- 
duct of  its  two  digits,  the  quotient  is  2,  and  if  27  is 
added  to  it  the  digits  are  reversed  ? 

2.  There  are  three  numbers,  the  difference  of  whose  dif- 
ferences is  8  ;  then-  sum  is  41  ;  and  the  sum  of  their 
squaft'es  is  699.     What  are  the  numbers  ? 

Notation. — Let  x  be  the  second  number,  and  y  the  difference  be- 
tween the  second  and  first,  so  that  x- —  y  represents  the  first.  Now  as 
the  difference  of  their  differences  is  8,  and  as  the  second  is  y  more  than 
the  first,  the  third  is  ?/  -f-  8  more  than  the  second,  and  hence  is  x  -f- 
2/  -f-  8.  The  first  equation  is  3x  -|-  8  =  41,  and  the  second  (11  —  y)'^ 
-j- 121  -f-  (19  -|-  y)'-^  =  699.  Having  solved  the  problem  in  this  manner 
let  the  student  solve  it  by  letting  x,  y,  and  z  represent  the  numbers,  and 
then  compare  the  solutions. 

3.  There  are  three  numbers,  the  difference  of  whose  dif- 
ferences is  5  ;  their  sum  is  44  ;  and  their  product  is 
1950.     What  are  the  numbers? 

4.  A  grocer  sold  80  lbs.  of  mace  and  100  lbs.  of  cloves  for 
$65  ;  but  he  sold  60  lbs.  more  of  cloves  for  $20,  than 
he  did  of  mace  for  $10.  "What  was  the  price  of  a  pound 
of  each? 

5.  A  and  p  have  each  a  small  field,  in  the  shape  of  an  ex- 
act square,  and  it  requires  200  rods  of  fence  to  enclose 
both.  The  contents  of  these  fields  are  1300  square 
rods.  What  is  the  value  of  each,  at  $2.25  per  square 
rod?  Ans.,  One,  $900  ;  other,  $2,025. 


6.  Find  two  numbers,  such  that  the  sum  of  their  squares 
being  subtracted  from  tliree  times  their  product,  11  re- 
main ;  and  the  difference  of  their  squares  being  sub- 
tracted from  twice  their  product,  the  remainder  is  14. 

SuG.^This  example  gives  rise  to  homogeneous  equations  (114:). 

7.  What  two  numbers  are  those  whose  difference  multi- 
phed  by  the  difference  of  their  squares  is  32,  and  whose 
sum  multiphed  by  the  sum  of  their  squares  is  272  ? 


376  SIMULTANEOUS    EQUATIONS 

8.  The  difference  of  two  numbers  is  2,  and  the  square  of 
their  quotient  added  to  four  times  their  quotient  is  9*. 
What  are  the  numbers?  Ans.,  5,  and  3. 

9.  There  are  two  numbers,  whose  sum  multipHed  by  the 
less,  is  equal  to  4  times  the  greater,  but  whose  sum 
multiplied  by  the  greater  is  equal  to  9  times  'the  less. 
What  are  numbers? 

10.  Find  two  numbers,  such,  that  their  product  added  to 
their  sum  shall  be  47,  and  their  sum  taken  from  the 
sum  of  their  squares  shall  leave  62.        Ans.,  5  and  7. 

11.  Find  two  numbers,  such,  that  their  sum,  their  product, 
and  the  difference  of  their  squares  shall  be  all  equal  to 
each  other.  Ans.,  |  -f  ^^5,  and  ^  -\-  ^v^S. 

12.  Find  two  numbers  whose  product  is  equal  to  the  dif- 
ference of  their  squares,  and  the  sum  of  their  squares 
equal  to  the  difference  of  their  cubes. 

Am.,  iv/5,  and  J(5  +  ^5). 

13.  A  person  has  $1,300,  which  he  divides  into  two  portions, 
and  loans  at  different  rates  of  interest,  so  that  the  two 
portions  produce  equal  returns.  If  the  first  portion 
had  been  loaned  at  the  second  rate  of  interest,  it  would 
have  produced  $36,  and  if  the  second  portion  had  been 
loaned  at  the  first  rate  of  interest,  it  would  have  pro- 
duced $49.     Eequired  the  rates  of  interest. 

Ans.,  7  and  6  per  cent. 

14.  The  fore  wheel  of  a  wagon  makes  6  revolutions  more 
than  the  hind  wheel  in  going  120  yards  ;  but  if  the 
periphery  of  each  wheel  be  increased  1  yard,  the  fore 
wheel  will  make  only  4  revolutions  more  than  the  hind 
wheel  in  going  the  same  distance.  What  is  the  circum- 
ference of  each  wheel  ?  Ans.,  4  and  5. 

15.  The  sum  of  two  numbers  is  8  and  the  sum  of  their 
cubes  is  152,  what  are  the  numbers? 


OF   THE   SECOND   DEGREE.  377 

SuG. — Let  X  -}-  y  he  one  of  the  numbers,  and  x  ■ —  y  the  other. 
Thena;  ==  4,  and  2^3  +6xy^  =  152  [111). 

16.  The  sum  of  two  numbers  is  7,  and  the  sum  of  their 
4th  powers  is  641.     What  are  the  numbers  ? 

17.  The  sum  of  two  numbers  is  6,  and  the  sum  of  their 
5th  powers  is  1056.     "What  are  the  numbers  ? 

18.  The  product  of  two  numbers  is  24,  and  their  sum  mul- 
tiphed  by  their  difference  is  20  ;  find  them. 

Ans.,  4  and  6. 

19.  What  two  numbers  are  those  whose  sum  multipHed  by 
the  greater  is  120,  and  whose  difference  multiphed  by 
the  less  is  16  ?  Ans.,  2  and  10. 

20.  What  two  numbers  are  those  whose  sum  added  to  the 
sum  of  their  squares  is  42,  and  whose  product  is  15  ? 

A)is.,  3  and  5. 

21.  A's  and  B's  shares  in  a  speculation  altogether  amount 
to  $500  ;  they  sell  out  at  pa7\  A  at  the  end  of  2  years, 
B  of  8,  and  each  receives  in  capital  and  profits  $297. 
How  much  did  each  embark  ? 

Ans.,  A,  $275 ;  B,  $225. 

SuG. — Letting  x  be  A's  capital,  and  y  B's,  A  gained  297  —  x,  and  B, 
297  T—  y.  And  as  the  gains  are  proportioned  to  the  products  of  the 
respective  times  into  the  capitals,  2x  :  8y  :  :  297  —  x  :  297  —  y. 


22.    What  three  numbers  are  those  in  A.  P.,  whose  sum  is 
120,  and  the  sum  of  whose  squares  is  5600  ? 

Ans.,  20,  40,  60. 

SuG. — La  solving  examples  involving  several  quantities  in  arithmet- 
ical progression,  it  is  usually  expedient  to  represent  the  middle  one  of 
the  series,  when  the  number  of  terms  is  odd,  by  x,  and  let  y  be  the 
common  difference.  If  the  number  of  terms  is  even,  represent  the  two 
middle  terms  by  x  —  y,  and  x  -^  y,  making  the  common  difference  2y. 
Thus  the  statement  of  the  above  problem  becomes  x  —  y  -\-  ^  -{z^'^hU 
r=  120,  or  3x  =  120  ;  and  {x  —  2/)^  +  ic^  -f  (x  -4-  vj^  =?  5600,  pr  3x^  -f 
2t/2=5600.  '^ 


378  SIMULTANEOUS   EQUATIONS 

23.  What  four  numbers  are  those  in  A.  P.,  the  sum  of 

whose  squares  is  84,  and  their  product  105  ? 

SuG, — Call  tbie  numbers  x  —  Sy,x  —  y,x  -\-  y,  and  x  +  3?/. 

24.  The  sum  of  five  numbers  in  A.  P.  is  35,  and  the  sum 
of  their  squares  285  ;  find  the  numbers. 

25.  What  three  numbers  are  those  in  A.  P.,  the  sum  of 
whose  squares  is  1232,  and  the  square  of  the  mean 
greater  than  the  product  of  the  two  extremes,  by  16  ? 

26.  Find  four  numbers  in  A.  P.  such,  that  the  sum  of  the 
squares  of  the  extremes  is  4500,  and  the  sum  of  the 
squares  of  the  means  is  4100. 

27.  The  product  of  five  numbers  in  A.  P.  is  945  ;  and  their 
sum  is  25.     What  are  the  numbers  ? 

Ans.,  1,  3,  5,  7,  9. 

28.  The  product  of  four  numbers  in  A.  P.  is  280,  and  the 
sum  of  their  squares  166  ;  find  them. 

29.  The  sum  of  nine  numbers  in  A.  P.  is  45,  and  the  sum 
of  their  squares  285  ;  find  them. 

Ans.,  1,  2,  3,  etc.,  to  9. 

30.  The  sum  of  seven  numbers  in  A.  P.  is  35,  and  the  sum 
of  their  cubes  1295  ;  find  them. 

Ans.,  2,  3,  etc.,  to  8.     i 


31.  There  are  three  numbers  in  geometrical  progression, 
whose  sum  is  52,  and  the  sum  of  the  extremes  is  to 
the  mean  as  10  to  3.     What  are  the  numbers? 

Ans.,  4,  12,  and  36. 

SuG.  — Let  X  be  the  first  term  and  y  the  ratio.  Then  x-}-xy-{-  xy^  =  52, 
and  x  +  xy^  :xy  ::10  :d,  orl-\-y^  :y  ::10  :  3. 

32.  The  sum  of  three  numbers  in  geometrical  progression 


OF   THE   SECOND   DEGREE.  379 

is  13,  and  the  product  of  the  mean  and  sum  of  the  ex- 
tremes is  30.     What  are  the  numbers  ? 

Ans.,  1,  3,  and  9. 

SuG. — We  have  x  -\- xij  -{-  xy^  =  13,  and  (x  -\-  xy-)  xy  =  30.     From 

30 
the  first  x  -f-  ^y'^  =  Vi  —  xy,   and  from  the  second,  a;  -f-  ^-  =  —  '■> 

xy 

30 
whence  13  —  xy=^  — ,  or  x-y-  —  13a:?/  =  —  30. 

33.  If  the  seventh  and  tenth  terms  of  a  geometrical  pro- 
gression are  6  and  750  respectively,  what  are  the  inter- 
mediate terms  ? 

SuG. — The  equations  area;?/6=:  6,  and  a'^/^  =  750.  Divide  the  second 
by  the  first. 

34.  If  the  third  and  fifth  terms  of  a  geometrical  j^rogTession 
be  75  and  300  respectively,  what  will  the  fourth  term 
be?  Ans.,  150. 

35.  If  the  first  and  fourth  terms  of  a  geometrical  progres- 
sion are  3  and  24  respectively,  what  are  the  two  inter- 
mediate terms? 

36.  There  are  four  numbers  in  geometrical  progi'ession. 
The  sum  of  the  means  is  30,  and  the  product  of  the 
extremes  200,  what  are  the  numbers  ? 

SuG. — Using  the  notation  given  above,  we  have  for  the  equations 
xy-^-xy^-z^  30,  and  x'^y^  =  200.     But  a  more  elegant  and  simple  solu- 

x^  v^ 

tion  is  obtained  by  representing  the  numbers  by  — ,  x,  y,  and  — ,  in 

y  X 

which  -  is  the  ratio.     This  gives  for  the  equations  x  -\-  y  =  SO,  and 

xy  =  200. 

When  this  form  of  notation  is  used  for  an  odd  number  of  terms,  it 
is  expedient  to  make  xy  the  middle  term.  Thus  for  five  terms,  we  have 
0-3      ^  y^ 

— ,  X-,  xy,  y-,  — . 

37.  The  sum  of  three  numbers  in  G.  P.  is  26,  and  the  sum 
of  their  squares  is  364  ;  required  the  numbers. 

SuG. — The  equations  are  x^  +  xj/  +  ^^  =  26,  and  x*  +  .r^yz  +  y*  =  364, 
which  have  already  been  solved. 


380 


SYNOPSIS. 


£  j  Quadratic  Equation  \  P^i^e. -Incomplete. 
o  i  ^  ^  i  Affected. — Complete. 


Root  of  Equation. 
Frob.    To  Solve.     Bern. 


Cor.  1.  Number  of  Boots.    Dem. 
Cor.  2.  Imaginary  Roots. 
r  Prob.  To  Solve.     Rule.     Bern. 

Completing  Square.    What  ?  |  <^^^-.  ^lethod. 

(  Special  methods. 
Cor.  1.  Roots  of  an  Aff.  Quad.     Dem. 
Cor.  2.  To  -write  the  root  of  x-  +  px  =  q  directly. 
Pure.     Prop.  1. 

r  Prop.  2.   Dem.     When  solved  as  Quad. 


eS 


in 

S 


j  Expedients.     ]  ^^^i^-  3-    Cor. 
[  (  Prop.  4:.    How  applied. 

Bern. 
Bern. 
Bern. 
Sch. 


Affected. 

Prop.  1. 

Prop.  2. 

Prop.  3. 

Prop.  4. 

Expedients.     Enumerate  the  7  given. 

r  Common  method. 

^  A.  P.  -{  j  Even  number  of  terms. 

!  Special   methods,  i  _  , , 
[    A  (Odd         "         "       " 

r  Common  method. 

A.  G.  P.  ]  j  Even  number  of  terms. 

Special  methods.  S  ^^^ 
I  (  Odd         "         "       " 


Test  Questions. — If  one  of  two  equations  between  two  unknown 
quantities  is  of  the  first  degree,  and  the  other  of  the  second,  what  will 
be  the  degree  of  the  resulting  equation  after  ehminating  one  of  the 
unknown  quantities?  Prove  it.  How  may  such  equations  be  solved? 
In  general,  what  is  the  degree  of  the  equation  arising  from  eliminating 
one  unknown  quantity  from  two  equations,  each  of  the  second  degree  ? 
Prove  it.  Mention  the  several  cases  g'ven  in  which  such  equations 
can  be  solved  by  quadratics,  and  state  the  process  in  eaclt  c^se. 


LOGARITHMS.  381 

CHAPTER  Y. 
logahithms. 


[Note. — It  is  the  purpose  of  this  chapter  to  give  a  simple.  arithmet« 
ical  view  of  the  nature  of  logarithms,  with  some  illustrations  of  their 
practical  utility.  For  the  production  of  the  Logarithmic  Series  and 
its  use  in  computing  Logarithms,  see  Appendix  II.  Enough,  how- 
ever, is  here  given  for  practical  use  in  trigonometry,  as  usually 
studied,  and  to  enable  the  student  to  understand  the  use  of  loga- 
rithms in  ordinary  operations.] 

117 »  A  Logarithm  is  the  exponent  by  which  a  fixed 
number  is  to  be  aJBfected  in  order  to  produce  any  required 
number.  The  fixed  number  is  called  the  Base  of  the 
System. 

III. — Let  the  Base  be  3  :  then  the  logarithm  of  9  is  2  ;  of  27,  3  ;  of 
81,  4;  of  19683,  9;  for  32  =  9;  3^=27;  3^  =  81;  and  33  =  19683. 

Again,  if  64  is  the  base,  the  logarithm  of  8  is  i,  or  .5,  since  64^,  or 
64-"' =  8  ;  {.  e.,  i,  or  .5  is  the  exponent  by  which  64,  the  base,  is  to  be 
affected  in  order  to  produce  the  number  8.     So  also ,  64  being  the  base, 

i,  or  .333  -f-  is  the  logarithm  of  4,  since  64* ,  or  64-333+  =  4  ;  i.  e.,  •^,  or 
.333  -}-  is  the  exponent  by  which  64,  the  base,  is  to  be  affected  in  order 

to  produce  the  number  4.     Once  more,  since  64^,  or  64-ece+  =  16,  §, 

_  i 
or  .666+  is  the  logarithm  of  16,  if  the  base  is  64.     Finally,  64     ^,  or 
64.  —  .5  __  i^  Qj.  j^25  ;   hence  —  i.  or  —  .5,  is  the  logarithm  of  g,  or  .125, 
when  the  base  is  64.     In  like  manner,  with  the  same  base,  — i,  or 
—  .333  -f-  is  the  logarithm  of  4,  or  .25. 

EXAMPLES. 

1.   If  2  is  the  base  what  is  the  logarithm  of  4  ?   of  8  ?   of 
82?  of  128?  of  1024? 

MODEL   SOLUTION. 

7  is  the  logarithm  of  128,  if  2  is  the  base,  since  7  is  the  exponent  by 
which  2  is  to  be  affected  in  order  to  produce  the  number  128. 


382  LOGARITHMS. 

2.  If  5  is  the  base  what  is  the  logarithm  of  625  ?  of  15625? 
of  125?  of  25? 

3.  If  10  is  the  base  what  is  the  logarithm  of  100?  of  1,000? 


of  10,000?  of  10,000,000? 

If  2  is  the  base  what  is  the 
-1-,  or  .125?  of  ^\,  or  .03125? 


4.  If  2  is  the  base  what  is  the  logarithm  of  ^,  or  .25  ?   cf 

Ans.  to  the  last,  —  5. 


5.  If  8  is  the  base,  of  what  number  is  f ,  or  .666  +  the 
logarithm?  of  what  number  is  f  or  1.333+  the  loga- 
rithm ?  of  what  number  is  2  the  logarithm  ?  of  what 
number  is  2^,  or  2.333  +  ?  of  what  number  3f,  or 
3.666  +  ?  Ans.  to  the  last,  2048. 

.  ScH. — Since  any  number  with  0  for  its  exponent  is  1,  the  logarithm 
of  1  is  0,  in  all  systems.  Thus  10^  =  1,  whence  0  is  the  logarithm 
of  1,  in  a  system  in  which  the  base  is  10. 

118,  A  St/stein  of  Logarithms  is  a  scheme  by 
which  all  numbers  can  be  represented,  either  exactly  or 
approximately,  by  exponents  by  which  a  fixed  number  (the 
base)  can  be  affected.  Negative  numbers  can  have  no 
logarithms. 

110,  There  are  Two  Systems  of  Logarithms  in  common 
use,  called,  respectively,  the  Briggeaiv  or  Common  System, 
and  the  Najnerian  or  Hyperbolic  System.  The  base  of  the 
former  is  10,  and  of  the  latter  2.71828  +. 


120 •  One  of  the  most  important  uses  of  logarithms  is 
to  facilitate  the  multiplication,  division,  involution,  and  the 
extraction  of  roots  of  large  numbers.  These  processes  are 
performed  upon  the  following  principles  : 


121,  JProp,  1,  TJie  sum  of  the  logarithms  of  two  num-^ 

hers  is  the  logarithm  of  their  p?vduct 


LOGAKITHMS.  •  38? 

Dem. — Let  a  be  the  base  of  the  system.  Let  m  and  n  be  any  two 
numbers  whose  logarithms  are  x  and  y  respectively.  Then  by  defini- 
tion a^  =  m,  and  a^  r=  n.  Multiplying  these  equations  together  we 
have  a'^+y  =  mn.     Whence  x-\-  y  is  the  logarithm  of  mn.     q.  e.  d. 

122.  ^rop,  2.  The  logarithm  of  the  quotient  of  two 
numbers  is  the  logarithm  of  the  dividend  minus  the  logarithm 
of  the  divisor. 

Dem. — Let  a  be  the  base  of  the  system,  and  m  and  n  any  two  num- 
bers whose  logarithms  are,  respectively,  a*,  and  y.     Then  by  definition 

we  have  ai^=m.  and  a"  =  n.     Dividing,  we  have  a*  — "  =  — .     "Whence 

n 

X — w  is  the  logarithm  of  — .     q.  e  .  d. 

71 


123.  J^rop.  3.  The  logarithm  of  a  jwwer  of  a  number 
is  the  logarithm  of  the  number  multiplied  bij  the  index  of  the 
power. 

Dem. — Let  a  be  the  base,  and  x  the  logarithm  of  m.  Then  a''^=^m', 
and  raising  both  to  any  power,  as  the  2th,  we  have  a^^  =  m~.  Whence 
xz  is  the  logarithm  of  the  zth  power  of  m.     q.  e.  d. 


124,  Prop.  4:.  The  logarithm  of  any  root  of  a  number 
is  the  logarithm  of  the  number  divided  by  the  number  express- 
ing the  degree  of  the  root. 

Dem. — Let  a  be  the  base,  9.nd  x  the  logarithm  of  m.     Then  a^=m. 

X 

Extracting  the  zth  root  we  have  a^  =  ym.     Whence  -   is   the    loga- 
rithm of  v/?u.     Q.  e.  d. 


12 o.  In  order  to  apply  these  principles  practically,  we 
need  what  is  called  a  Table  of  Logarithms.  That  is,  a  table 
from  which  we  can  readily  obtain  the  logarithm  of  any 
number,  or  the  number  corresponding  to  any  logarithm. 
We  will,  therefore,  proceed  to  show  how  such  a  table  can 


384  -  LOGARITHMS. 

be  computed.  Though  the  method  about  to  be  given  is 
not  the  most  expeditious  now  known,  it  is,  nevertheless,  the 
one  used  when  our  tables  were  first  computed. 


jL20»  JProb,  To  compute  the  common  logarithm,  of  any 
decimal  number. 

Dem. — 1st.  It  is  evident  that  it  is  only  necessary  to  compute  the  log- 
arithms of  prime  numbers,  since  the  logarithm  of  a  composite  number 
is  the  sum  of  the  logarithms  of  its  factors  {121). 

2nd.  To  compute  the  logarithms  of  the  series  of  prime  numbers. 
In  the  first  place  we  know  that  the  logarithm  of  1  is  0,  since  10^  =  1. 
Also  the  logarithm  of  10  is  1,  since  10'  =  10.  Now  if  we  find  the  log- 
arithm of  5,  we  can  get  the  logarithm  of  2  by  subtracting  the  logarithm 
of  5  from  log.  10.  *  If  there  be  any  number  which  is  the  logarithm  of 
5  it  is  evident  it  must  lie  between  0,  which  is  log.  1,  and  1,  which  is  log. 
10.     Therefore  starting  with       10^  =  1     and 

10'  =10  multiplying  them  together 
we  have  10'  =  lO 

Extracting  the  square  root,  10-5  =  v/lO  =  3.162277  +. 
Again,  as  5  lies  between  10  and  3 .162277  +  its  logarithm  lies  between  1 
and  .5,     Multiplying  the  last  two  equations  we  have  10' '5  =  31. 62277+. 
Extracting  the  square  root,  10-75=  v/^i.  ^-A'zn  +  =  5. 623413  -f-. 

Again,  10-     =3.162277  4- 

and  10-"   =5.623413  4- 

Multiplying  10'--  =  17.78^7895914  4- 

Extracting  square  root  lO-csa  =  v/17 .7827895914  -f  =  4.2169644- 

Again,  multiplying  this  last  by  10 -^^  =:  5.623413  -{-,  as  5  lies  between 
these  numbers,  and  extracting  the  square  root,  we  have  10  •^'^■'■5==: 
4.869674  4--  I^  each  case  the  exponent  of  10  is  the  logarithm  of  the 
number  ;  thus  .6875  is  log.  4.869674-)-.  Continuing  this  process  to  22 
operations  (!)  we  have  log.  5. 000000 -|-=  . 698970 -{-,  which  i3  suffi- 
ciently accurate  for  ordinary  purposes. 

Now  log.  10  —  log.  5  =  log.  2  =  1  —  .698970  =  .301030. 

To  find  log.  3,  we  would  take  lO-^  =  3.162277  4-,  and  10°  =  l,and 
proceed  as  before. 

Were  it  our  purpose  to  find  log.  11,  the  computation  would  be  as 
follows  : 

*  This  is  the  common  abbreviation  of  "logarithm  of  10,"  and  should  be  read 
"logarithm  of  10,"  not  "  log  ten,"  which  is  grossly  inelegant. 


LOGABITHMS.  385 

101  =  10 

102  ^  100 
10^  =  1000 

10-  =  101-5  =  v/lOoi)  =  31.62277  -f 
101  =  10 


10^ 

101 

101 

=  102-5  =  316.2277  + 
•25=^316.2277+  = 

17.78278  + 
10 

10^ 

101 
lOi 

••Zb  = 
•  125  - 

-.  177.827«  + 

=  v/l77.8278  +  == 

13.33521  + 
10 

102 
101 
101 

.125  = 
•0625  : 

=  133.  3521  + 

=  v/l33.3621  +  = 

= 

=  11.54782  — 
10 

102 

101 
101 

•0025  : 
•03125 
•0625  : 

=  115.4782  + 

=  >/ll5.4782  +  = 

=  10.74607  + 
11.54782  — 

102- 09375 
101.04687; 
101-03125 

=  124.09368  + 
*  =  V'124. 09358  + 

.  =  11.13973  + 
10.74607  + 

102.07S12O    =  119.70845  + 

101  •0390625  =  10.94113  + 

Whence  1.0390625  is  log.  10.94113;  and  proceeding  with  the  computa- 
tion, the  logarithm  of  11  may  be  found  -with  suflacient  accuracy. 

Example. — Let  the  pupil  compute  the  logarithm  of  23, 
and  compare  his  result  with  the  logarithm  as  found  in  the 
table  following. 

ScH. — The  pupil  will  not  fail  to  be  impressed  with  an  idea  of  the  im- 
mense labor  involved  in  computing  a  tabic  of  logarithms.  The  com- 
mon tables  give  the  logarithms  of  numbers  from  1  to  10,000,  with 
provision,  as  mil  be  seen  hereafter,  for  using  them  to  find  the  loga- 
rithms of  much  larger  numbers,  with  sufficient  accuracy  for  practical 
purposes.     One  page  of  such  a  table  is  given.    (Page  386.) 


127*  JProb,    To  find  the  logarithm  of  a  number  from 
\he  table. 


386 

(a  page  of)  a 

TABLE  OF  LOGARITHMS. 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

I*. 

280 

447158 

7313 

7468 

7623 

7778 

7933 

8088 

8242 

8397 

8552 

155 

281 

8706 

8861 

9015 

9170 

9324 

9478 

9633 

9787 

9941 

••95 

154 

282 

450249 

0403 

0557 

0711 

0865 

1018 

1172 

1320 

1479 

1633 

154 

283 

1786 

1940 

2093 

2247 

2400 

2553 

2700 

2859 

3012 

3165 

153 

281 

3318 

3471 

3624 

3777 

3930 

4082 

4235 

4387 

4540 

4692 

153 

285 

4845 

4997 

5150 

5302 

5454 

5606 

5758 

5910 

6062 

6214 

152 

286 

6366 

6518 

6670 

6821 

6973 

7125 

7276 

7428 

7579 

7731 

152 

287 

7882 

8033 

8184 

8330 

8487 

8638 

8789 

8940 

9091 

9242 

151 

288 

9392 

9543 

9694 

9845 

9995 

•146 

•296 

•447 

•597 

•748 

151 

289 

460898 

1048 

1198 

1348 

1499 

1649 

1799 

1948 

2098 

2248 

150 

290 

2398 

2548 

2697 

2847 

2997 

3146 

3296 

3445 

3594 

3744 

150 

291 

3893 

4042 

4191 

4340 

4490 

4639 

4788 

4936 

5085 

5234 

149 

292 

5383 

5532 

5080 

5829 

5977 

6126 

6274 

6423 

6571 

6719 

149 

293 

6868 

7016 

7164 

7312 

7460 

7608 

7756 

7904 

8052 

8200 

148 

294 

8347 

8495 

8643 

8790 

8938 

9085 

9233 

9380 

9527 

9675 

148 

295 

9822 

9969 

•116 

•263 

•410 

•557 

•704 

•851 

•998 

1145 

147 

296 

471292 

1438 

1585 

1732 

1878 

2025 

2171 

2318 

2464 

2610 

146 

297 

2756 

2903 

3049 

3195 

3341 

3487 

3633 

3779 

3925 

4071 

146 

298 

4216 

4362 

4508 

4653 

4799 

4944 

5090 

5235 

5381 

5526 

146 

299 

5671 

5816 

5962 

6107 

6252 

6397 

6542 

6687 

6832 

6976 

145 

300 

7121 

7266 

7411 

7555 

7700 

7844 

7989 

8133 

8278 

8422 

145 

301 

8566 

8711 

8855 

8999 

9143 

9287 

9431 

9575 

9719 

9863 

144 

302 

480007 

0151 

0294 

0438 

0582 

0725 

0869 

1012 

1156 

1299 

144 

303 

1443 

1586 

1729 

1872 

2016 

2159 

2302 

2445 

2588 

2731 

143 

304 

2874 

3016 

3159 

3302 

3445 

3587 

3730 

3872 

4015 

4157 

143 

305 

4300 

4442 

4585 

4727 

4869 

5011 

5153 

5295 

5437 

5579 

142 

306 

5721 

5863 

6005 

6147 

6289 

6430 

6572 

6714 

6855 

6997 

142 

307 

7138 

7280 

7421 

7563 

7704 

7845 

7986 

8127 

8269 

8410 

141 

308 

8551 

8692 

8833 

8974 

9114 

9255 

9396 

9537 

9677 

9818 

141 

309 

9958 

-99 

•239 

•380 

•520 

•661 

•801 

•941 

1081 

1222 

140 

310 

491362 

1502 

1642 

1782 

1922 

2062 

2201 

2341 

2481 

2621 

140 

311 

2760 

2900 

3040 

3179 

3319 

3453 

3597 

3737 

3876 

4015 

139 

312 

4155 

4294 

4433 

4572 

4711 

4850 

4989 

5128 

5267 

5406 

139 

313 

5544 

5683 

5822 

5960 

6099 

6238 

6376 

6515 

6653 

6791 

139 

314 

6930 

7063 

7206 

7344 

7483 

7621 

7759 

7897 

8035 

8173 

138 

315 

8311 

8148 

8586 

8724 

8862 

8999 

9137 

9275 

9412 

9550 

138 

316 

9687 

9824 

9962 

-99 

•236 

•374 

•511 

•S48 

•785 

•922 

137 

317 

501059 

1196 

1333 

1470 

1607 

1744 

1880 

2017 

2154 

2291 

137 

318 

2427 

2564 

2700 

2837 

2973 

3109 

3246 

3382 

3518 

3655 

136 

319 

3791 

3927 

4063 

4199 

4335 

4471 

4607 

4743 

4878 

5014 

136 

320 

5150 

5286 

5421 

5557 

5693 

5828 

5964 

6099 

6234 

6370 

136 

321 

6505 

6640 

6776 

6911 

7046 

7181 

7316 

7451 

7588 

7721 

135 

322 

7856 

7991 

8126 

8260 

8395 

8530 

8664 

8799 

8934 

9068 

135 

323 

9203 

9337 

9471 

9006 

9740 

9874 

•••9 

•143 

•277 

•411 

134 

324 

510545 

0679 

0813 

0947 

1081 

1215 

1349 

1482 

1616 

1750 

134 

325 

1883 

2017 

2151 

2284 

2418 

2551 

2684 

2818 

2951 

3084 

133 

326, 

3218 

3351 

3484 

3617 

3750 

3883 

4010 

4149 

4282 

4414 

133 

327 

4548 

4681 

4813 

4946 

5079 

5211 

5344 

5476 

5609 

5741 

133 

328 

8874 

6006 

6139 

6271 

6403 

6535 

6608 

6800 

6932 

7064 

132 

329 

7196 

7328 

7460 

7592 

7724 

7855 

7987 

8119 

8251 

8382 

132 

330 

N. 

8514 

8646 

8777 

8909 

9040 

9171 

9303 

9434 

9566 

9697 

131 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

LOGARITHMS.  387 

Solution. — Page  386  is  one  page  of  a  table  of  logarithms  giving  the 
logarithms  of  numbers  from  1  to  10,000  directly,  and  from  which  the 
logarithms  of  other  numbers  can  also  be  found  with  little  trouble. 
Thus,  let  it  be  required  to  find  the  logarithm  of  325.  Now,  as  the  loga- 
rithm of  100  is  2,  and  of  1000  is  3,  the  logarithm  of  325  must  be  between 
2  and  3,  i  e.  2  and  a  fraction.  The  fractional  part  is  all  that  is  given 
in  the  table,  as  the  integral  can  be  kno-«Ti  by  simple  inspection.  Look- 
ing in  the  table  down  the  column  marked  N  (numbers),  we  find  325, ^ 
and  opposite  it  in  the  column  headed  0,  we  find  1883,  but  just  above 
this  we  observe  51 ,  which  belongs  to  this  logarithm  and  which  is  simply 
omitted  to  save  space  in  the  table,  since  it  really  belongs  as  a  prefix  to 
all  the  logarithms  clear  do-vvTi  to  the  number  332  where  it  is  replaced 
by  52.  Prefixing  the  51  to  the  1883,  we  have  .511883  as  the  decimal 
part  of  the  logarithm.  Hence  log.  325  is  2.511883.  In  like  manner  the 
logarithm  of  any  number  consisting  of  three  figures  is  found  from  the 
table. 

To  find  the  logarithm  of  a  number  consisting  of  four  figures.  Let  it  be 
required  to  find  the  logarithm  of  2936.  Looking  for  293  (the  first  three 
figures)  in  the  column  of  numbers,  and  then  passing  to  the  right  until 
reaching  the  column  headed  6,  the  fourth  figure,  we  find  7756,  to  which 
prefixing  the  figures  46,  which  belong  to  all  the  logarithms  following 
them  till  some  others  are  indicated,  we  have  for  the  decimal  part  of  the 
logarithm  of  2936,  .467756.  But,  as  3  is  the  logarithm  of  1000,  and  4  of 
10,000,  log.  2936  is  3  and  this  decimal,  or  log.  2936  =  3.467756. 

To  find  the  logarithm  of  a  number  consisting  of  more  than  4  figures. 
Let  it  be  required  to  find  the  logarithm  of  2845672.  Finding  the  deci- 
mal part  of  logarithm  of  the  first  4  figures  2845,  as  before,  we  find  it 
to  be  .454082.  Now  the  logarithm  of  2846  is  153  (millionths,  really) 
more  than  that  of  2845.  Hence,  assuming  that  if  an  increase  of  the 
number  by  1000  makes  an  increase  in  its  logarithm  of  153,  an  increase 
of  672  in  the  number,  wiUmake  an  increase  in  the  logarithm  of  -]^o~i.ro  o^ 
.672  of  153,  or  103,  omitting  lower  orders,  and  adding  this  to  .454082, 
we  have  .454185  as  the  decimal  part  of  log.  2845672.  The  integral  part 
is  6,  since  2845672  lies  between  the  6th  and  7th  powers  of  10.  Hence 
log.  284672  =  6.454185.     q.  e.  d. 

ScH.  1. — If  in  seeking  the  logarithm  of  any  number  any  of  the  heavy 
dots  noticed  in  the  table  are  passed,  their  places  are  to  be  filled  with 
O's,  and  the  first  two  figures  of  the  decimal  of  the  logarithm  taken  from 
the  0  column  in  the  line  below.  Thus  log.  3166  is  3.500511.  This  ar- 
rangement of  the  table  is  a  mere  matter  of  convenience  to  save  space. 


388  LOGARITHMS. 

ScH.  2, — The  column  marked  D  is  called  the  column  of  Tabular 
Differences  ;  and  any  number  in  it  is  the  difference  between  the  loga- 
rithms found  in  columns  4  and  5,  which  is  usually  the  same  as  be- 
tween any  two  consecutive  logarithms  in  the  same  horizontal  line.  The 
assumption  made  in  using  this  difference  ;  viz.,  that  the  logarithms 
increase  in  the  same  ratio  as  the  numbers ,  is  only  approximately  true, 
but  still  is  accurate  enough  for  ordinary  use. 

128^  The  Integral  Part  of  a  logarithm  is  called  the 
Characteristic^  and  the  decimal  part  the  Man^ 
tissa. 

129.  I*rop,  The  Mantissa  of  a  decimal  fraction,  or  of 
a  mixed  number,  is  the  same  as  the  mantissa  of  the  number 
considered  as  integral. 

Dem.— Above  it  was  found  that  log.  2845672=6.454185.  Now  this 
means  that  lO^-* « 4 1 8  s  =:=  2845672.  Dividing  by  10  successively  we  have 

105.464185  ^  284567.2,           or  log.  284567.2           =  5.454185, 

lQ4.464i85^    28456.72,         or  log.  28456.72         =4.454185, 

103.464186^     2845.672,       or  log.  2845.672        =3.454185, 

102-464185^       284.5672,     or  log.  284.5672      =2.454185, 

101.454186^         28.45672,    or  log.  28.45672    =1.454185, 

100-464186  _           2.845672,  or  log.  2.845672  =  0.454185. 

Now  if  we  continue  the  operation  of  division,  only  writing  0. 454185 —  X 
1,454185,  meaning  by  this  that  the  characteristic  is  negative  and  the 
mantissa  positive,  and  the  subtraction  not  performed,  we  have 

10T.4o4i8.5  _  .2845672,       or  log.  .2845672      —1.454185, 

10iT454i85  =- .02845672,     or  log.  .02845672    =2.454185, 

IOT.454185  =  .002845672,  or  log.  .002845672=3.454185, 
etc.,  etc.     Q.  E.  D. 

ScH. — The  characteristic  of  an  integral  number,  or  of  a  mixed  integral 
number  and  decimal,  is  one  less  than  the  number  of  integral  places, 
as  will  appear  by  comparing  such  numbers  with  the  powers  of  10,  as  is 
done  in  demonstrating  {12S).  The  characteristic  of  a  number  en- 
tirely decimal  fractional,  is  negative,  and  one  greater  than  the  number 
«of  O's  immediately  followinier  the  decimal  point,  as  appears  from  the  last 
demonstration,  or  as  appears  from  the  fact  that  10~^  =  xV  =  .1: 
10-2  =  T^o  =  .01 ;  10-2  ^  ^i__  ^  ^001 ;  etc.,  etc. 


LOGAKITHMS.  389 

EXAMPLES. 

Find  the  logarithms  of  the  following  numbers  :  285  ;  3145  ; 
2905624  ;  30942716  ;  298.026  ;  32.5614 ;  2.8641 ;  .3205  ; 
.00317  ;  00000328. 
Results,  log.  298.026  =  2.474254  ;  log.  .00317  =  3.501059. 


130,  PtoJ),  —  To  find  a  number  corresponding  to  a 
given  logarithm. 

Solution. — Let  it  be  required  lo  find  the  number  corresponding  to 
tbe  logarithm  5.515264.  Looking  in  the  table  for  the  next  less  mantissa, 
we  find  .515211,  the  number  corresponding  to  which  is  3275  (no  account 
now  being  taken  as  to  whether  it  is  integral,  fractional  or  mixed  ;  as 
in  any  case  the  figures  will  be  the  same).  Now,  from  the  tabular  differ- 
ence, in  column  D,  we  find  that  an  increase  of  133  (millionths,  really) 
upon  this  logarithm  (.515211),  vrould  make  an  increase  of  1  in  the 
number,  making  it  3276.  But  the  given  logarithm  is  only  53  greater 
than  this,  hence  it  is  assumed  (though  only  approximately  correct) 
that  the  increase  of  the  number  is  -^^  of  1,  or  53  -j- 133  =  .398-4  +. 
This  added  (the  figures  annexed)  to  3275,  gives  32753984+.  The 
characteristic,  being  5,  indicates  that  the  number  hes  between  the  5th 
and  6th  powers  of  10,  and  hence  has  6  integral  places.  .  • .  -5.515264 
=  log.  327539.84+. 

EXAMPLES. 

Find  the  numbers  corresponding  to  the  following  loga- 
rithms :  3.467521;  2.467521;  0.467521;  4.500281; 
1.520281;  4.520281;  0.52081;  1.520281;  2.490160; 
2.490160;  and  0.490160. 

Results,  2.490160  =  log.  309.1435  +  ;   2.490160  =  log. 
.030914  +  ;  0.490160  =  log.  3.091435 +. 


131,  As  logarithms  are  largely  used  to  facilitate  numer- 
ical computations,  it  is  important  that  the  student  be  able 
to  take  any  formula  representing  such  operations  and  write 
at  once  the  equivalent  logarithmic  operations. 


390  LOGARITHMS. 

EXAMPLES. 

1.  If  28.035  :  3.2781  :  :  3114.27  :  x,  what  logarithmic  op- 
erations will  find  X  ? 

SvG. — The  logarithm  of  the  product  of  the  means  is  the  sum  of 
their  logarithms  ;  and  the  logarithm  of  the  quotient  of  this  product 
divided  by  the  first  extreme,  is  the  logarithm  of  said  product  minus  the 
logarithm  of  the  other  extreme.  .  • .  log.iC  =  log.  3. 2781  -}- log.  3114.27 
—  log.  28.035  =0.515622  +  3.493356—1.447700  =  2.561278.  Having  a 
table  sufficiently  extended,  the  number  corresponding  to  this  logarithm 
could  be  found,  and  would  be  the  value  of  x. 

2.  Find  the  product  of  23.14,  by  5.062,  knowing  that  log. 
23.14  is  1.364363,  log.  5.062  is  0.704322,  and  log. 
117.1347  is  2.068685. 

3.  How  is  287  raised  to  the  5th  power  by  means  of  loga- 
rithms ?     How  is  the  5th  root  extracted  ? 

4.  Extract  the  5th  root  of  31152784.1  by  means  of  loga- 
rithms. 

SuG.— Log.  >y31 152784.1  =  i  log.  31152784.1  =  1.498699.  The  num- 
ber, therefore,  is  31.52  -J-. 

5.  What  is  the  cube  root  of  30?  Ans.,  3.107  +. 

6.  What  is  the  cube  root  of  .03  ? 

SuG.— Log.  .03  =  2.477121.  Now  to  divide  this  by  3,  we  have  to 
bear  in  mind  that  the  characteristic  alone  is  negative,  i.  e.,  2.477121 
=— 2-1-.477121,  or— 1.522879.  This  divided  by  3  gives  —  .507626, 
or  0  — .  507626  =1. 492374.  But  a  more  convenient  method  of  effecting 
this  division  is  to  write  for  the  — 2,  — 3-f-l,  whence  we  have  foi 
2.477121,  —3  4-1.477121,  which  divided  by  3  gives~l. 492374,  nearly. 

7.  Divide  3.261453  by  2,  by  4,  by  5. 

Last  quotient,  1.4522906. 


APPENDIX 


SECTION    I. 
DIFFERENTIATION. 

[This  subject  is  inserted  as  the  best  metliod  of  reaching  the 
demonstration  of  the  Binomial  Formula  and  the  production  of  the 
Logarithmic  Series.  While  it  is  equally  simple,  to  say  the  least, 
■with  the  old  method,  it  is  more  direct,  and  gives  the  student  nothing 
but  what  is  of  fundamental  importance  in  subsequent  mathematical 
work.] 

132,  In  certain  classes  of  problems  and  discussions 
the  quantities  involved  are  distinguished  as  Constant  and 
Variable. 

133,  A  Constant  quantity  is  one  which  maintains 
the  same  value  throughout  the  same  discussion,  and  is 
represented  in  the  notation  by  one  of  the  leading  letters 
of  the  alphabet. 

134,  Variable  quantities  are  such  as  may  assume  in 
the  same  discussion  any  value  within  certain  limits  deter- 
mined by  the  nature  of  the  problem,  and  are  represented 
by  the  final  letters  of  the  alphabet. 

III. — If  X  is  the  radius  of  a  circle  and  y  is  its  area,  y  =  -nx^,  as  we 
learn  from  Geometry,  tt  being  about  3.1416.     Now  if  x,  the  radius, 

*  The  preceding  part  of  this  volume  famishes  a  course  in  Algebra  quite  as  full 
as  will  be  found  practicable  or  desirable  in  most  high  schools  and  academies,  and  is 
an  adequate  preparation  for  college.  This  appendix,  selected  from  Olnet's  Uni- 
versity Algebra,  is  inserted  for  such  of  tht-  above  schools  as  desire  a  fuller  conrse, 
and  as  adapting  the  book  to  the  needs  of  many  of  our  colleges  which  do  not  tod 
it  expedient  to  give  as  much  time  to  this  subject  as  is  required  to  master  the 
University  Algebra.  There  is  nothing  in  the  ordinary  college  course  wuiclx 
requireB  more  Algebra  than  is  found  in  this  volume. 


392  DIFFERENTIATION. 

varies,  y,  the  area,  will  vary  ;  but  tt  remains  tlie  same  for  all  values 
of  a;  and  y.  In  this  case  x  and  y  are  the  variables,  and  it  is 
a  constant. 

Again,  if  y  is  the  distance  a  body  falls  in  time  x,  it  is  evident 
that  the  greater  x  is,  the  greater  is  y,  i.  e.,  that  as  x  varies  y  varies. 
We  learn  from  Physics  that  y  =  16  j^^^.  for  comparatively  small 
distances  above  the  surface  of  the  earth.  In  the  expression  y  = 
lQ^^a^\  X  and  y  are  the  variables,  and  IGyi^  is  a  constant. 

Once  more,  suppose  we  have  y^  —  25iC^— 3a;'^— 5,  as  an  expressed 
relation  between  x  and  y,  and  that  this  is  the  only  relation  which  is 
required  to  exist  between  them  ;  it  is  evident  that  we  may  give 
values  to  x  at  pleasure,  and  thus  obtain  corresponding  values  for  y. 
Thus  if  2!  =  1,  2^  =  ±  -v/l^»  \ix  =  2,y=  ±/v/l83,  etc.,  etc.  In  such  a 
case  X  and  y  are  called  variables.  But  we  notice  that  if  we  give  to  x 
such  a  value  as  to  make  3a;2  +  5>25a;3  (as,  for  example,  |,  \,  etc.),  y 
will  be  imaginary.  This  is  the  kind  of  limitation  referred  to  in  our 
definition  of  variables. 

135,  ScH. — The  pupil  needs  to  guard  against  the  notion  that  the 
terms  constant  and  variable  are  synonyms  for  known  and  unknown, 
and  the  more  so  as  the  notation  might  lead  him  into  this  error.  The 
quantities  he  has  been  accustomed  to  consider  in  Arithmetic  and 
Elementary  Algebra  have  all  been  constant.  The  distinction  here 
made  is  a  new  one  to  him,  and  pertains  to  a  new  class  of  problems 
and  discussions. 

136,  A  Function  is  a  quantity,  or  a  mathematical 
expression,  conceived  as  depending  for  its  yalue  upon  some 
other  quantity  or  quantities. 

III. — A  man's  wages  for  a  given  time  is  a  function  of  the  amount 
received  per  day,  or,  in  general,  his  wages  is  a  function  of  the  time 
he  works  and  the  amount  he  receives  per  day.  In  the  expression 
y  =  1Qy2X^  {134),  second  illustration,  y  is  a  function  of  x,  i.  e.,  the 
space  fallen  through  is  a  function  of  the  time.  The  expression 
2ax^—dx-\-5b,  or  any  expression  containing  x,  may  be  spoken  of  as  a 
function  of  x. 

J37,  When  wc  wish  to  indicate  that  one  variable,  as  y, 
is  a  function  of  another,  as  x,  and  do  not  care  to  be  more 
specific,  we  write  y=f{x)y  and  read  "?/  equals  (or  is)  a 


DIFFEREI^-TIATION.  393 

function  of  a:."  This  means  nothing  more  than  that  y  is 
equal  to  some  expression  containing  the  variable  x,  and 
which  may  contain  any  constants.  If  we  wish  to  indicate 
several  different  expressions  each  of  which  contains  x,  we 
write/(a;),  go  [x),  or/'  (re),  etc.,  and  read  "  the/ function  of 
a:,"  "  the  (f>  function  of  x"  or  "the/'  function  of  x." 

III. — The  expression /(a?)  may  stand  for  a^— 2a? +  5,  or  for  Z{a^—x'^), 
or  for  any  expression  containing  x  combined  in  any  way  with  itself  or 
with  constants.  But  in  the  same  discussion  f{x)  will  mean  the  same 
thing  throughout.  So  again,  if  in  a  particular  discussion  we  have  a 
certain  expression  containing  x{e.  g.,  dx^—ax  +  2ah).  it  may  be  repre- 
sented by /(a-),  while  some  other  function  of  x,  e.  g.,  5  {a^—x^)  +  2x'^, 
might  be  represented  by/'  (x),  or  (p  (x). 

138.  In  equations  expressing  the  relation  betweeen  two 
variables,  as  in  y"^  =  daa^—x^,  it  is  customary  to  speak  of 
one  of  the  variables,  as  y,  as  a  function  of  the  other,  x. 
Moreover,  it  is  convenient  to  think  of  x  as  varying  and 
thus  producing  change  in  y.  When  so  considered,  x  is 
called  the  Independent  and  y  the  Dependeyit  variable.  Or 
we  may  speak  of «/  as  a  function  of  the  variable  x. 

139.  An  Infinitesimal  is  a  quantity  conceived 
under  such  a  form,  or  law,  as  to  be  necessarily  less  than 
any  assignable  quantity. 

Infinitesimals  are  the  increments  by  which  continuous 
number,  or  quantity  (8),  may  be  conceived  to  change  value, 
or  grow. 

III. — Time  affords  a  good  illustration  of  continuous  quantity,  or 
number.  Thus  a  period  of  time,  as  5  hours,  increases,  or  grows,  to 
another  period,  as  7  hours,  by  infinitesimal  increments,  i.  e.,  not  by 
hours,  minutes,  or  even  seconds,  but  by  elements  which  are  less  than 
any  assignable  quantity. 

140.  Consecutive  Values  of  a  variable  are  values 
which  differ  from  each  other  by  less  than  any  assignable 
quantity,  i.  e.,  by  an  infinitesimal.    Consecutive  values  of  a 


394  DIFFERENTIATIOlf. 

function  are  values  which  correspond  to  consecutive  values 
of  its  variable. 

141.  A  Differeiitictl  of  a  function,  or  variable,  is  the 
difference  between  two  consecutive  states  of  the  function, 
or  variable.     It  is  the  same  as  an  infinitesimal. 

III. — Resuming  the  illustration  y  —  IQ^^x^  (134),  let  x  be  thought 
of  as  some  particular  period  of  time  (as  5  seconds),  and  y  as  the  dis- 
tance through  which  the  body  falls  in  that  time.  Also,  let  x'  represent 
a  period  of  time  infinitesimally  greater  than  x,  and  y'  the  distance 
through  which  the  body  falls  in  time  x'.  Then  x  and  x'  are  consecu- 
tive values  of  x,  and  y  and  y'  are  consecutive  values  of  y.  Again,  the 
difference  between  x  and  of,  as  x'—x,  is  a  differential  of  the  variable 
X,  and  y'  —y  is  a  differential  of  the  function  y. 

142.  Notation. — A  differential  of  x  is  represented  by 
writing  the  letter  d  before  x,  thus  dx.  Also,  dy  means,  and 
is  read  "differential  2/." 

Caution. — Do  not  read  dx  by  naming  the  letters  as  you  do  ax ;  but 
read  it  "  differential  x."  The  d  is  not  a  factor,  but  an  abbreviation  for 
the  word  differential. 

143.  To  Differentiate  a  function  is  to  find  an 
expression  for  the  increment  of  the  function  due  to  an 
infinitesimal  increment  of  the  variable ;  or  it  is  the  process 
of  finding  the  relation  between  the  infinitesimal  increment 
of  the  variable  and  the  corresponding  increment  of  tlie 
function. 

RULES    FOR    DIFFERENTIATING. 

144.  Rule  I. — To  differentiate  a  single  valuable, 
simply  write  the  letter  d  before  it. 

This  is  merely  doing  what  the  notation  requires.  Thus,  if  x  and 
x'  are  consecutive  states  of  the  variable  x,  i.  e.,  if  x'  is  what  x  becomes 
when  it  has  taken  an  infinitesimal  increment,  x'—x  is  the  differential 
of  X,  and  is  to  be  written  dx.  In  like  manner,  y'  —y  is  to  be  written 
dy,  y'  and  y  being  consecutive  values. 


DIFFEREIJTIATIOK.  395 

145,  Rule  U, — Constant  factors  or  divisors  ap- 
pear in  the  differential  the  same  as  in  the  function. 

Dem. — Let  us  take  tlie  function  y  =  ax,  in  wliich  a  is  any  constant, 
integral  or  fractional.  Let  x  take  an  infinitesimal  increment  dx, 
becoming  x  +  dx;  and  let  dy  be  the  corresponding  *  increment  of  y,  so 
that  when  x  becomes  x  +  dx^  y  becomes  y  +  dy.     We  then  have 

1st  state  of  the  function     .    .  y  =:  ax; 

2d,  or  consecutive  state     .     .    y  +  dy  =  a  {x+dx)  =  ax+adx. 

Subtracting  the  1st  from  the  2d  dy  =  adx, 

which  result  being  the  diflference  between  two  consecutive  states  of 
the  function,  is  its  differential  {14:1).  Now  a  appears  in  the  differ- 
ential just  as  it  was  in  the  function.     This  would  evidently  be  the 

same  if  a  were  a  fraction,  as  — .  We  should  then  have,  in  like  man- 
ner,  dy  =  —dx2^  the  differential  ofy=—x.    o.  e.  d. 


146.  Rule  m. — Constant  terms  disappear  in  dif- 
ferentiating ;  or  the  differential  of  a  constant  is  0. 

Dem. — Let  us  take  the  function  y  =  ax  +  h,  in.  which  a  and  h  are 
constant.  Let  x  take  an  infinitesimal  increment  and  become  x  +  dx  ; 
and  let  dy  be  the  increment  which  y  takes  in  consequence  of  this 
change  in  x,  so  that  when  x  becomes  x+dx,  y  becomes  y  +  dy.  We 
then  have 

1st  state  of  the  function     .    .  y  =  ax  +  h; 

2d,  or  consecutive  state      .    .    y  +  dy  =  a{x+dx)  +  b  =  ax+ ado  +  h. 

Subtracting  the  1st  from  the  2d  dy  =  adx, 

which  being  the  difference  between  two  consecutive  states  of  the 
function,  is  its  differential  {141).  Now  from  this  differential  the 
constant  b  has  disappeared. 

We  may  also  say  that  as  a  constant  retains  the  same  value,  there 
is  no  difference  between  its  consecutive  states  (properly  it  has  no  con- 
secutive states).  Hence  the  differential  of  a  constant  may  be  spoken 
of  (though  with  some  latitude)  as  0.     Q,  E,  d. 

*  The  word  "  contemporaneous  "  is  often  used  in  this  connection. 


396  DIFFERENT!  ATIOlir. 

14:7,  Rule  rV.  —  To  differentiate  the  algebraic 
sum  of  several  variables,  diff^ereutlate  each  term  sep- 
arately and  connect  the  differentials  with  the  same 
signs  as  the  terms. 

Dem. — Let  u  =  x  +  y—2,  u  representing  the  algebraic  sum  of  the 
variables  x,  y,  and  —z.  Then  is  du  =  dx  +  dy—dz.  For  let  dx,  dy, 
and  dz  be  infinitesimal  increments  of  x,  y,  and  z  ;  and  let  du  be  the 
increment  which  u  takes  in  consequence  of  the  infinitesimal  changes 
in  X,  y,  and  z.    We  then  have 

1st  state  of  the  function.     .     *     .  u  =  x-vy—z  ; 

2d,  or  consecutive  state  .     .     .     .    u  +  du  =  x  +  dx  +  y  +  dy—{z  +  dz). 

Or u  +  du  =  X  +  dx  +  y  +  dy—z—dz. 

Subtracting  the  1st  state  from  the  2d      du  =  dx+dy—dz.    Q.  e.  d. 


148.  Rule  V.  —  Tlie  differential  of  the  product 
of  two  variables  is  the  differential  of  the  first  into 
the  second,  plus  the  differential  of  the  second  into  the 
first. 

Dem. — Let  u  ■=  xy  be  the  first  state  of  the  function.  The  consecu- 
tive state  is  u-\-dii  =  {x  +  dx){y+ dy)  —  xy+ ydx + xdy  +  dx ■  dy.  Sub- 
tracting  the  1st  state  from  the  consecutive  state  we  have  the  difler- 
ential,  i.  e.,du  =  ydx  +  xdy  +  dx  •  dy.  But,  as  dx  •  dy  is  the  product  of 
two  infinitesimals,  it  is  infinitely  less  than  the  other  terms  {ydx  and 
xdy),  and  hence,  having  no  value  as  compared  with  them,  is  to  be 
dropped.*    Therefore,  du  =  ydx  +  xdy.    Q.  E.  D. 

*  It  will  doubtless  appear  to  the  pupil,  at  first,  as  if  this  j^ave  a  result  only 
approximately  correct.  Such  is  not  the  fact.  The  result  is  absolutely  correct.  Ko 
error  is  introduced  hy  dropping  dx  ■  dy.  In  fact  this  term  must  be  dropped  accord- 
ins:  to  the  nature  of  infinitesimals.  Notice  that  by  definition  a  quantity  which  is 
infinitesimal  with  respect  to  another  is  one  which  has  no  assic^able  magnitude 
•with  reference  to  that  other.  Hence  we  must  so  treat  it  in  our  reasoning.  Now 
dxdy  is  an  infinitesimal  of  an  infinitesimal  (i.e.,  two  infinitesimals  multiplied 
together),  and  hence  is  infinitesimal  with  reference  to  ydx  and  xdy,  and  must  ba 
treated  as  having  no  assignable  value  with  respect  to  them;  that  is,  it  must  be 
dropped. 


DIFFEKEKTIATIOK.  397 

149.  Rule  VI. — TJie  differential  of  the  product 
of  several  variables  is  the  sum  of  the  products  of  the 
differential  of  each  into  the  product  of  all  the 
others. 

Dem. — Let  u  =  xyz;  then  du  ■=  yzdx  +  a:zdy-\-xydz.  For  the  1st 
state  of  the  function  is  w  =  xyz,  and  the  2d,  or  consecutive  state, 
u  +  du  =  (x  +  dx)  iy^dy)  iz  +  dz),  or  u-\-du  =  xyz-iryzdx  +  xzdy^-xydz  + 
xdydz+ydxdz+zdxdy  +  dxdydz.  Subtracting,  and  dropping  all  infini- 
tesimals of  infinitesimals  (see  preceding  rule  and  foot-note),  we  have 
du  =  yzdx  +  xzdy  +  xydz. 

In  a  similar  manner  the  rule  can  be  demonstrated  for  any  number 
of  variables.     Q.  E.  d. 


ISO.  Rule  VII. — TJie  differential  of  a  fraction 
having  a  variable  numerator  and  denominator  is  the 
differential  of  the  numerator  multiplied  by  the 
denominator,  minus  the  differential  of  the  denom- 
inator multiplied  by  the  numerator,  divided  by  the 
square  of  the  denoTninator. 

Dem. — Let  u  =  -;  then  is  du  =  ~ -„ — - .     For,  clearing  of  frac- 

tions,  yu  =  x.    Differentiating  this  by  Rule  5th,  we  have  udy+ydu  = 

X  xdv 

dx.      Substituting  for  u  its  value  -,  this  becomes  — ~  +  ydu  =  dx. 

y  y 

Finding  the  value  of  du,  we  have  du  =  - — —^ — -.    q.  e.  d. 

lol.  Cor. — Tlie  differential  of  a  fraction  having  a  con- 
stant numerator  and  a  variaUe  denominator  is  the  jjroduct 
of  the  numerator  ivith  its  sign  changed  into  the  differential 
f  the  denominator,  divided  by  the  square  of  the  denom- 
inator. 

Let  u  =  -.    Differentiating  this  by  the  rule  and  calling  the  dif- 
n 


398  DIFFEEEN^TIATION. 

f erential  of    the  constant  {a)  0,  we    have  du  = -^  = f- . 

Q.   E.   D. 

152,    SCH. — If  the  numerator  is  variable  and  the   denominator 
constant,  it  falls  under  Rule  2. 


158.  Rule  VIII.— I7ie  differential  of  a  variable 
affected  with  an  exponent  is  the  continued  product 
of  the  exponent,  the  variahle  with  its  exponent 
diminished  by  1,  and  the  diff^erential  of  the  variable. 

Dem.— 1st.  When  the  exponent  is  a  positive  integer.  Let  y  =  x"", 
m  being  a  positive  integer ;  then  dy—mx"'-^dx.  For  y=x'^=x  .x.x  .x. 
to  m  factors.  Now,  differentiating  this  by  Rule  6,  we  have  dy  = 
(xxx  .  .  to  m—1  factors)  dx+{xxx  .  .  to  m— 1  factors)  <Za;  +  etc.,  to  m 
terms;  or  dy  =  x"'-^dx  +  x"'-^dx  +  x'"-^dx+  etc.,  to  m  terms.  There- 
fore dy  =  mx'^-^dx. 

m 

2d.  When  tJie  exponent  is  a  positive  fraction.     Let  y  =  x"  ,  ^  being 

a  positive  fraction  ;  then  dy  =  -a;"     dx.    For  involving  both  members 
to  the  nth.  power  we  have  y^  =  x"^.   Differentiating  this  as  just  shown, 

we  have   ny^-'^dy  =  mx'^~^dx.     Now  from  y  =  x''  we  have  y"~'  = 

mn—m  rnn — m 

X    »    .     Substituting  this  in  the  last  it  becomes  nx    »    dy^mx'^-^dx ; 

Hin  —m  m 

whence  dy  =  '-^a;  "    c?a;  =  ^a;«     dx.    Q.  e.  d. 

3d.  When  the  exponent  is  negative.  Let  y  =  a;-",  n  being  integral 
or  fractional ;  then  dy  =  —nx-"~'^dx.     For  y  =  ar-«  =  — ,  which  dif- 

ferentiated  by   Rule  7,  Cor.,  gives   dy  = —  =  —  nx-^'-^dx, 

Q.  E.  D. 


EXAM  P  LES. 

1.  Differentiate  y  =  3x^—2x-\-4:. 

Solution.— The  result  is  dy  =  6xdx  —  2dx.     Which  is  thus 


DIFFEKEKTIATION.  399 

obtained  :  By  Rule  1,  the  differential  of  y  is  dy.  To  differentiate  the 
second  member  we  differentiate  each  term  separately  according  to 
Rule  4.  .  In  differentiating  3^;-,  we  observe  that  the  factor  3  is 
retained  in  the  differential,  Rule  3,  and  the  differential  of  x'^  is,  by 
Rule  8,  %xdx.  Hence,  the  differential  of  3a;-  is  ^xdx.  The  differential 
of  —2a!  is  —%dx.  By  Rule  3,  the  constant  4  disappears  from  the  dif- 
ferential, or  its  differential  is  0. 

2.  Differentiate  y  ■=.  2ax^-\-4:aa:^—x-{-m. 

Result,  dy  =  AiCixdx -\-12ax\lx — dx. 

3.  Differentiate  y  =  5bx^—30x^-{-4.x. 

4.  Differentiate  y  =  Ax^-\-Bx^-\-  Cx\ 

154.  SCH. — It  is  desirable  that  the  pupil  not  only  become  expert 
in  writing  out  the  differentials  of  such  expressions  as  the  above,  but 
that  he  know  what  the  operation  signifies.  Thus,  suppose  we  have 
the  equation  y  =  5x.  This  expresses  a  relation  between  x  and  y. 
Now,  if  X  changes  value,  y  must  change  also  in  order  to  keep  the 
equation  true.  In  this  simple  case  it  is  easy  to  see  that  y  must 
change  5  times  as  fast  as  x  in  order  to  keep  the  equation  true. 
This  is  what  differentiation  shows.  Thus,  differentiating,  we  have 
dy  =  Mx.  That  is,  if  x  takes  an  infinitesimal  increment,  y  takes  an 
infinitesimal  increment  equal  to  5  times  that  which  x  takes  ;  or,  in 
other  words,  y  increases  5  times  as  fast  as  x. 

Now  let  us  take  a  case  which  is  not  so  simple.  Let  2/=Sa;'^— 2a;  +  4, 
and  let  it  be  required  to  find  the  relative  rate  of  change  of  x  and  y. 
Differentiating,  we  have  dy  —  6xdx—2dx  =  (6a;— 2)  dx.  This  shows 
that,  if  X  takes  an  infinitesimal  increment  represented  by  dx,  y  takes 
one  (represented  by  dy)  which  is  6a;— 2  times  as  large  ;  i.  e.,  that  y 
increases  6a;— 2  times  as  fast  as  x.  Notice  that  in  this  case  the  rela- 
tive rate  of  increase  of  x  and  y  depends  on  the  value  of  x.  Thus, 
when  x=l,  y  is  increasing  4  times  as  fast  as  x  ;  when  x=2,  y  is 
increasing  10  times  as  fast  as  x ;  when  x  =  S,  y  is  increasing  16  times 
as  fast  as  x ;  etc. 

5.  Differentiate  y  =  x^ — x^,  and  explain  the  significance 
of  the  result  as  above.  Result,  dy  =z  {5x^—3x^)  dx. 

6.  In  order  to  keep  the  relation  2y  =  Sx^  true  as  x  varies, 
how  must  y  vary  in  relation  to  ic  ?    What  is  the  relative 


400  DIFFERENTIATION". 

rate  of  change  when  x  =  4:?    When  cc  =  27    When  x  =  l? 

V(henx=z  i?     When.T  =  |? 

Ansivers.  When  x  =  4:,  y  increases  12  times  as  fast  as  x. 

When  X  =  ^,  y  increases  at  the  same  rate  as  x.    In  general 

y  increases  Sx  times  as  fast  as  x.     When  x  is  less  than  -J-,  y 

increases  slower  than  x. 

2x^  x^ 1 

7  to  12.  Differentiate  the  following :  u  =  7-— ;  u  =  ~ : 

^  Sy  '  a:2+l' 

y  =  x'^z^;    u  =  x^y^-\-6x;    y  z=z  x^—dx^-^4:X^—x^-{-l;    and 
y  =  ^x^—^x^-{-x. 

13  to  17.    Differentiate    y  =  (d^^x^y;     y  =  {a  +  x^)^ ; 

y  z=  (3a;— 2)*;  y  =  (2— a:2)-2;  and  y  =  {l-^xf^. 

Sug's, — Such  examples  should  be  solved  by  considering  the  entire 
quantity  within  the  parenthesis  as  the  variable.  This  is  evidently 
admissible,  since  any  expression  which  contains  a  variable  is  variable 
when  taken  as  a  whole.  Thus  to  differentiate  y  =  {a  +  x^)s,  we  take 
the  continued  product  of  the  exponent  (f),  the  variable  {a  +  x'^)  with 
its  exponent  diminished  by  1,  [i.  e.,  (a  +  x'^fi],  and  the  differential  of 
the  variable  (i.  e.  the  differential  of  a  +  x-,  which  is  2xdx).     This  gives 

us  dy  =  i{a  +  x'')~hxdx,  or  dy  =  ix{a  +  x^)~hx  =  _J^^ 


18  to  22.    Differentiate 


dy^a  +  x^ 
1 


1+x'    (l+a;)2'     (l+xy 


1  ;i  1 

—7n — :    and    — m 


{l-\-xy'  (l  +  a:)3 

23.  In  the  expression  6x^,  when  x  is  greater  than  1  does 
the  function  {6x^)  change  faster  or  slower  than  x?  How, 
when  X  is  less  than  ^  ?  What  does  the  process  of  differ' 
entiating  6x^  signify  ? 

Answer  to  the  last.  Finding  the  relative  rate  of  change  of  6a^ 
and  X,  or  finding  what  increment  6;^^  takes  when  x  takes  the 
increment  dx. 


INDETERMINATE    COEFFICIEKTS.  401 

Or,  in  still  other  words,  finding  the  difference  between  two  con- 
secutive states  of  Qx^,  and  hence  the  relation  between  an  infinitesimal 
increment  of  x  and  the  corresponding  increment  of  6^^. 


INDETERMINATE     COEFFICIENTS. 

ISS,  Indeterminate  Coefficients  are  coefficients 
assumed  in  the  demonstration  of  a  theorem  or  the  solution 
of  a  problem,  wliose  values  are  not  known  at  the  outset,  but 
are  to  be  determined  by  subsequent  processes. 


loO.  Prop.— If  A-f  Bx  +  Cx2  +  Dx^  +  etc.  =  A! -\- 
B'x  +  C'x=^+D'x3-f-  etc.,  in  ivMcli  x  is  a  variaUe  and  the 
coefficients  A,  B,  A'  B',  etc.,  are  constants,  tlie  coefficients  of 
the  like  poioers  of -s.  are  equal  to  each  other.  That  is,  K  =^ 
A'  {these  heing  the  coefficients  ofx^),  B  =  B',  C  =  C,  etc. 

Dem. — Since  the  equation  is  true  for  any  value  of  x,  it  is  true  for 
x=  0.  Substituting  this  value,  we  have  A  =  A'.  Now  as  A  =  A' 
are  constant,  thej  have  the  same  values  whatever  the  value  assigned 
to  X.  Hence  for  any  value  of  a;,  ^  =  A'.  Again,  dropping  A  and  xi', 
we  have  Bx+Cx^  +  Dx^+  etc.  =  B'x  +  C'x^  +  I>'a^+  etc.,  which  is  true 
for  any  value  of  x.  Dividing  by  x,  we  obtain  B+Cx  +  Dx^+  etc.  = 
B'  +  Ox  +  D'x'^  +  etc.,  likewise  true  for  any  value  of  x.  Making  a;  —  0, 
B  ■=  B',  as  before.  In  this  manner  we  may  proceed,  and  show  that 
0  =C',I)  =  B',  etc.     Q.  E.  D. 

1S7.  Con.— If  A  +  Bx-|-Cx2  +  Dx3-f-  etc.  =  0,  is  true 
for  all  values  ofx,  each  of  the  coefficients  A,  B,  C,  etc.,  is  0. 

For  we  may  write  A  +  Bx  +  Cx^  +  Dx^  +  Ex^  +  Fx""  +  etc.  =  0  +  0.r  + 
0a;2  4-0.r3-f-0a;^-l-0x=+  etc.  Whence  by  the  proposition  ^  =  0,  B  =  Q, 
C=0,  etc. 


402        DEMONSTRATION   OF  THE   BINOMIAL  FORMULA. 

SECTION     II. 

DEMONSTRATION  OF  THE  BINOMIAL  FORMULA. 

lo8.  Theorem, — Letting  x  and  y  represent  any  quan- 
tities tvhatever  (i.  e.  he  variables)  and  m  a7i2/  coiistant, 

/    ,     n™         «  .       ml.   mim—l)    „,  „  „     m(m—l){m  —  2) 

Dem. — We  may  write  {x  +  y)""  ==  x'"\\+    \  .    Now  put  -  =  z  and 
assume 

(l  +  z)"'  =  A  +  Bz+  Cz''  +  I)z^  +  Ez*  +  Fz'>  +  etc.,  (1) 

in  which  A,  B,  C,  etc.,  are  indeterminate  coefficients  independent  of 
z  [i.  e.  constants),  and  are  to  be  determined.  To  determine  these  co- 
efficients we  proceed  as  follows  : 

Differentiating  (1),  we  have 

m{\+  zy-Hz  =  Bdzi'2Czdz  +  ZDzHz  +  4J]]z'clz  +  ^Fz^dz  +  etc. 

Dividing  by  dz,  we  have 

m{\  4-  z^-'  =  B  +  2Cz  +  SBz^  +  4:Ez^  +  5Fz*+  etc.  (3) 

Differentiating  (2)  and  dividing  by  dz,  we  have 

m(7?j-l)(l+2)™-2  =2(7+3  •  dBz  +  d  -  4Ez' +  4: .  5Fz^  +  etc.    (3) 

Differentiating  (3)  and  dividing  by  dz,  we  have 
w(m- l)(m-2)(l  +  2)"^-3  =  2  .  3i)  +  2  .  3  .  4^s  +  3  .  4 .  5M^  +  etc.    (4) 

Differentiating  (4)  and  dividing  by  dz,  we  have 
m(m-l){m-2){m-d)[l  +2)'"-^  =  2.3.  4^+2  .3.4.  5^2+  etc.    (5) 

Differentiating  (5)  and  dividing  by  dz,  we  have 

m{m-l)(m-2){m-d)(m-i){l  +  z)"'-'  =  2.S.4:.5F+  etc.        (6) 


*  This  form  is  read  "  factorial  S,"  "  factorial  4,"  etc. ;  and  signifies  the  product 
of  the  natural  numbers  from  1  to  3, 1  to  4,  etc. 


DEMONSTRATION^   OF  THE   BINOMIAL  THEOREM.        403 

We  liave  now  gone  far  enough  to  enable  us  to  determine  the  co- 
^  efficients  A,  B,  G,  B,  E,  and  F,  and  doubtless  to  determine  the  law  of 
the  series. 

As  all  the  above  equations  are  to  be  true  for  all  values  of  2,  and  as 
the  coefficients  A,  B,  G,  etc.,  are  constants,  i.  e.,  have  the  same  values 
for  one  value  of  z  as  for  another,  if  we  can  determine  their  values  for 
one  value  of  z,  these  will  be  their  values  in  all  cases.  Now,  making 
2  =  0,  we  have   from   (I)   J.  =  1 ;  from  {^),B  —  m;  from  (3),  (7  = 

— ^  - — -  (the  factor  1  being  introduced  into  the  denominator  for  the 

Bake  of  symmetry) ;  from  (4),  B  =  — ^  J^^~~^ ;    from  (5),  E  = 

m{m-l)(m-2)(m-S)     .         ...    ^     m(m-l)(m-2)(m-Sym-4) 
\i '       "^  ^^'  ^^ j5 • 

These  values  substituted  in  (1)  give 

m{m—l)        'm(m—l)(m—2)  .. 


(1+2)"'=  l+mz  + 


12  '  [3 


m{m—l)(m—2)(m—S)  ^      m{m-l){m-2)(m-d){m-A)  ^ 

+  rj  2    + ._-  2    +  etc. 

|4  |5 


Finally,  replacing  2  by  its  value  - ,  we  have 

X 

I A    y\^        S  -.       y    ^(^  —  1)  y^ 

{x  +  yY—a^\\  +  -\    =a;"M  1+w- + -^.3 ^—, 

\         X/  \  X  I  w  X' 

m(m  —  l)im-2)  y^      m(m—l)(m  —  2)(m-S)  1/ 

"^  ^  x^""  |4  x^ 

m{m  —  1)  (77*  —  2)  {m  —  3)  (m  —  A)y^ 
■^  [5  ¥  +  ^*^- 

,       m(m  —  1)  -     w(77i  —  1)  (m  —  2)        ,  . 

l£  E 

^  m{m-l){m-2){m-Z)  ^^_,^ 

^  m{m-l){m-2)  (m  -  8)  (m  -  4)  ^^^^^  ^  ^^^ 

159,  CoR.  1. — TJie  nth,  or  general  term  of  the  scries  is 
m(m-l)(m-2) (^-^  +  2)  ^_+,  _., 

"T3i  *^    y  • 


404  THE   LOGAKITHMIC   SERIES. 

For  we  observe  that  tlie  last  factor  in  tlie  numerator  of  tlie  co- 
efficient of  any  particular  term  is  m  —  the  number  of  the  term  less 
2,  i.  e.,  for  the  ?ith  term,  m  —  {71  —  2),  or  m  —  n  +  2;  and  the  last 
factor  in  the  denominator  is  the  number  of  the  term  —  1,  i.  e.,  for  the 
nth  term,  n—  1.  The  exponent  of  x  in  any  particular  term  is  m  — 
the  number  of  the  term  less  1,  i.  e.,  for  the  nth.  term,  m  —  (n  —  1),  or 
m  —  n  +  1;  and  the  exponent  of  y  in  any  term  is  one  less  than  the 
number  of  the  term,  i.  e.,  for  the  71th.  term,  n  —  1. 

160,  Def. — In  a  series  the  Scale  of  Relation  is  the 

relation  which  exists  between  any  term  or  set  of  terms  and 
the  next  term  or  set  of  terms. 

161,  Cor.  2. — 77ie  scale  of  relation  in  the  hinomial 

series  is  I ^  )  ^'  ^^^^^  t^^  ^i^  ^^^^^^  muUiplied  by 

this  produces  the  (n  +  l)th  term. 

This  is  readily  seen  by  inspecting  the  series,  or  by  writing  the 
(n  +  l)th  term  and  dividing  it  by  the  nth.  Thus,  substituting  in  the 
general  term  as  given  above,  n  +  1  for  n,  we  have 

in{m  —  1)  (m  —  2)  -----  (m  —  w  -i-  1) 

Iji  . 

as  the  (n  -f-  l)th  term.     This  divided  by  the  nth,  or  preceding  term, 
—  n  +  ly        /m  +  1 


n         X         \     n  J 


gives 

X 


SECT  ION    III. 

THE    LOGARITHMIC    SERIES. 

162,  The  Modulus  of  a  system  of  logarithms  is  a 
constant  factor  which  depends  upon  the  base  of  the  system 
and  characterizes  the  system. 

163*   ^rop, — The  differential  of  the  logarithm  of  a 


THE    LOGARITHMIC    SERIES.  405 

number  is  the  differential  of  the  numler  multiplied  by  the 
onodulus  of  the  system,  divided  by  the  number  ;  ' 

Or,  in  the  Napierian  system,  the  modulus  being  1,  the 
differefitial  of  the  logarithm  of  a  number  is  the  diff'erential 
of  the  number  divided  by  the  number. 

Dem. — Let  X  represent  any  number,  i.  e.  he  a  variable,  and  nhe  a, 
constant  sucb  that  y  =  x\  Then  log  y  =  n  log  x  {123).  Differen- 
tiating y  =  nf;  we  liave  dy  =  nx'^~'^dx ;  whence 


x^'-^dx      x^  ^        «/ ,       dx'  W 

—  dx      -dx     — 

Again,  whatever  the  differentials  of  log  y  and  log  x  are,  n  being  a 
constant  factor  we  shall  have  the  differential  of  log  y  equal  to  11  times 
the  differential  of  log  x,  which  may  be  written 

d  (log  y)  =  ,i'd  (log  x),  whence  n  =  j^~r^        (3.) 

Now  equating  the  values  of  n  as  represented  in  (1)  and  (2),  we  have 
dy 

T?r- — T  =  I"  •    Whence  d^(logw)  bears  the  same  ratio  to  — ,  as 
fZ(loga;)       dx  v    &  i//  ^' 

X 

d  (log  x)  does  to  —  .    Let  m  be  this  ratio.    Then  d  (log  y)  = ,  and 

X  y 

,■      .       mdx 
d{logx)  =  -^-. 

We  are  now  to  show  that  m  is  constant  and  depends  on  the  base  of 
the  system. 

To  do  this,  take  y  =  zn' ,  from  which  we  can  find  as  above  n'  = 


,,.  ^  ,  =  —■ .  Now  as  m  is  the  ratio  of  d  (log  v)  to  —  ,  it  is  also  the 
d(log  z)       dz  \    ^  if/       ^  ^ 

2 

ratio  of  d{\og  z)  to  —  ;  and  d(\og  z)  =  — -  ,     Thus  we  see  that  in  any 

z  z 

case  in  the  same  system  tbe  same  ratio  exists  between  the  differen- 
tial of  the  logaritlim  of  a  number  and  the  differential  of  the  number 
divided  by  the  number.    Therefore  wz  is  a  constant  factor. 


406  THE    LOGARITHMIC    SERIES. 

Aerain  —  =  w  -  indicates  the  relative  rate  of  change  of  a  loga- 
^        dx  X 

ritlim  and  its  number.     Now  it  is  evident  that  the  larger  the  base  the 

slower  the  logarithm  will  change  with  reference  to  the  number.    (See 

examples  under  Art  117.)    But  the  factor  -  varies  inversely  in  the 

number  ;  hence  m  must  vary  with,  or  be  a  function  of  the  base.* 


.7(>4.  Frob. — To  produce  the  logarithmic  series. 

Solution. — The  logarithmic  series,  which  is  the  foundation  of  th^ 
usual  method  of  computing  logarithms,  and  of  much  of  the  theory  of 
logarithms,  is  the  development  of  log  {1+x).  To  develop  log  (l  +  x), 
assume 

log{l  +  x)  =  A  +  Bx+Cx^  +  I)x^  +  m!^  +  Fx^+  etc.,  (1) 

in  which  a;  is  a  variable,  and  A,  B,  C,  etc.,  are  constants. 

Differentiating  (1),  we  have 

?^  =  Bdx  +  2Gxdx + ZDx'dx  +  4Mi?dx + hFx'^dx  +  etc. 

1  +  a; 

Dividing  by  dx, 

^     =  B+2Ckc+^Di^+4^xf'-^^F^+  etc.  (2) 

1+x 

Differentiating  (2),  and  dividing  by  dx,  we  have 

-m— J— -=2C+3.3Da;+3.4JS^2^.4.5j«7Tj.3+  etc.  (3) 

(1  +  xf 

Differentiating  (3),  and  dividing  by  2  and  by  dx,  we  have 

m ^ — rr3D  +  3.4£'ic  +  2-3.5i^a;2  +  'etc.  (4) 

Differentiating  (4),  and  dividing  by  3  and  dx,  we  have 

~m  - ^ —  =  4F+  4  •  5Fx  +  etc.  (5) 

(1  +  a;)4 

*  What  the  relation  of  the  modulus  to  the  base  is,  we  are  not  now  concerned  to 
know  ;  it  will  be  determined  hereafter. 

t  The  number  is  1+a; ;  hence  the  differential  is  m  times  the  differential  of  l+a? 
divided  by  the  number  1+x. 


THE    LOGAEITHMIC     SEKIES.  407 

Differentiating  (5),  and  dividing  by  4  and  dx,  we  have 

m  ^— ,  =  5i^+  etc  *  (6) 

(1  +  xf 

We  have  now  gone  far  enough  to  enable  us  to  determine  the  coef- 
ficients A,  B,  C,  D,  E,  and  F,  and  these  will  probably  reveal  the  law 
of  the  series. 

As  all  the  above  equations  are  to  be  true  for  all  values  of  x,  and  as 
the  coefficients  A,  B,  C,  etc.,  are  constant,  i.  e.,  have  the  same  values 
for  one  value  of  x  as  for  another,  if  we  can  determine  their  values  for 
one  value  of  x,  these  will  be  their  values  in  all  cases.  Now,  making 
aj  =  0,  we  have,  from  (1),  ^  =  log  1  =  0  ;  from  (2),  B  =  ni;  from  (3), 
G  =  -im ;  from  (4),  I)  =  lm;  from  (5),  E  =  —\m ;  from  (6),  F=\m. 
These  values  substituted  in  (1)  give 

/jj2  rr&  qA  nA 

log  (1  +  ic)  =  m  (^  -  -  +  -  -  ^  +  ^  -  etc), 

the  law  of  which  is  evident.    This  is  the  Logarithmic  Series,  and 
should  be  fixed  in  memory. 

ScH. — The  Napierian  system  of  logarithms  is  characterized  by  the 
modulus  being  l{m=  1).    Hence  the  Napierian  logarithmic  series  is 

,,       ^  x^     xfi     x^     a?        ^ 

log(l  +  a")  =  ir--  + --^  +  --  etc. 

165,  Cor.  1. — Tlie  logarithms  of  tlie  same  oiumler  171 
different  systems  are  to  each  other  as  the  moduli  of  those 
systems. 

This  is  evident  from  the  general  logarithmic  series.  Thus  the 
logarithm  of  1  +  a;  in  a  system  whose  modulus  is  m,  is  expressed 

/>^2  /pZ         /jA         /*>5 

log™  {\  +  x)\  =  m{x-^^  +  --'^  +  '--  etc.) ; 

and  the  logarithm  of  the  same  number  in  a  system  whose  modulus  is 
m!  is  expressed 

*  Of  course  the  student  will  o'bserve  what  forms  the  succeeding  terms  in  this 
and  the  other  similar  cases  would  have.  Thus  here  we  should  have  bF+bQGx  + 
3 -5. 75a;' +  etc. 

t  The  subscripts  m  and  m'  are  used  to  distinguish  between  the  systems,  as  log 
(1  +  cc)  is  not  the  same  in  one  system  as  in  the  other.  Read  logm  (1  +«),  "  logarithm 
of  1  +aj  in  a  system  whose  modulus  is  m,"  etc. 


408  THE    LOGARITHMIC    SERIES. 


flj^        0^       iC^        0^ 

log^/(l  +  2;)  =  m'(a;--  +  --^  +  --  etc.). 

Now,  as  the  number  (1  +  x)  is,  by  hypothesis,  the  same  in  both  cases, 
X  is  the  same.     Hence,  dividing  one  equation  by  the  other,  we  have 

\o^m  (1  +  a;)  _  m 
log,„/(l  ■\-x)~  m' ' 

166,  Cor.  2. — Havifig  the  logarithm  of  a  nwnber  in  the 
Napierian  system,  we  have  hut  to  multi2Jly  it  by  the  modulus 
of  any  other  system  to  obtain  the  logarithm  of  the  same  nu?n- 
her  in  the  latter  system. 

Or,  the  logarithm  of  a  number  in  any  system  divided  by 
the  logarithm  of  the  same  number  in  the  Napierian  system, 
gives  the  modulus  of  the  former  system. 

167.  JProb, — To  ado2ot  the  Napierian  logarithmic  series 
to  numerical  computation  so  that  it  can  be  conveniently  used 
for  computing  the  logarithms  of  numbers, 

/y»2  /vi3  /v»4  /Tfl5 

Dem. — That  log  (1  +a?)  =  a!  —  ^  +  ^ —  7"  +  ^+  ®tc.,  is  not  in  a 

.4        o        4        o 

practicable  form  for  computing  the  logarithms  of  numbers  will  be 

evident  if  we  make  the  attempt.     Thus,  suppose  we  wish  to  compute 

the  logarithm  of  3.     Making  X  —  2,  we  have  log  (1  +2)  =  log  3  =  2— 

22      2^      2*     2^  ,     ,      , 

—  +  5 —  7"  +  ^ —  ®*^*  ^  series  m  which  the  terms  are  growing 

^        o        4        O 

larger  and  larger  (a  diverging  series). 

We  wish  a  series  in  which  the  terms  will  grow  smaller  as  we 
extend  it  (a  converging  series).  Then  the  farther  we  extend  the 
series,  the  more  nearly  shall  we  approximate  the  logarithm  sought. 
To  obtain  such  a  series,  substitute  —x  for  x  in  the  Napierian  loga- 
rithmic series,  and  we  have 

x^     x^     x^     x^ 
log{l-x)  =  -x---------eto. 

Subtracting  this  from  the  former  series,  we  have 

log  (l  +  a;)-log  {1—x)  =  log  (r^)  =  %  {x  +  l^?  +  \x^  +  ^x''  -\-  etc.). 


THE    LOGAEITHMIC    SEKIES. 


409 


Now  put  X  =  - — -.  ,  whence  1  +  x 


2z  +  l      22  +  1 


23 


22 


^  ,  and   1±^  =  l±f .    Hence,  as  log  (—-]  =  log  (1  +  s)  -  log  g, 
4-1  1— a;        z  \   z  / 

substituting,  and  transposing, 

log(l  +  .)=Iog.  +  2(^j  +  g^3  +  5^,  +  ^^, 


+  etc, 


)(A) 


This  series  converges  quite  rapidly,  especially  for  large  values  of 
s,  and  is  convenient  for  use  in  computing  logarithms. 

168.  I^rob. — To  compute  the  Napierian  logarithms  of 
the  natural  mmiiers  1,  2,  3,  4,  etc.,  ad  liiitum. 

Solution. — In  the  first  place  we  remark  that  it  is  only  necessary 
to  compute  the  logarithms  of  prime  numbers,  since  the  logarithm  of 
a  composite  number  is  equal  to  the  sum  of  the  logarithms  of  its 
factors  (121). 

Therefore  beginning  with  1,  we  know  that  log  1=0  (ScH.,  p.  382). 

To  compute  the  logarithm  of  2,  make  s  =  1,  in  series  (A),  and  we 

havelog(l  +  l)-logl=log2  =  2(l  +  -^-3+J-,  +  J:-  +  -'3,+ 

1  1  1 


+ 


+  etc. 


11.811      13.313      15.81^ 

The  numerical  operations  are  conveniently  performed  as  follows : 


3 


2.00000000 


.66666667 

1 

.07407407 

3 

.00828045 

5 

.00091449 

7 

.00010161 

9 

.00001129 

11 

.00000125 

13 

.00000014 

15 

•.  log  2 


.66666667* 

.02469136 

.00164609 

.00013064 

.00001129 

.00000103 

.00000009 

.00000001 

.69314718  * 


*  Though  the  decimal  part  of  a  logarithm  is  generally  not  exact,  it  is  not  cus- 
tomary to  annex  the  +  sign. 


410 


THE    LOGAKITHMIC    SERIES. 


Second.    To  find  log  3,  make  z  =  2,  -whence 

log  S  =  log  3  +  3  (1  +  g^j.3  +  5^5  +  Jg,  +  Jg5  +  etc.). 


COMPUTATION. 

5 

2.00000000 

25 

.40000000 

1 

.40000000 

25 

.01600000 

3 

.00533333 

25 

.00064000 

5 

.00012800 

25 

.00002560 

7 

.00000366 

.00000102 

9 

.00000011 

.40546510 

log  2  =  .69314718 

.-.  log  3  = 

1.09861228 

Third.    To  find  log  4. 

Log  4  =  2  log  2  =  2  X  .69314718  =  1.38629436. 

FouKTH.     To  find  log  5.    Let  cc  =  4,  whence 

log     5     =      log    4   +   3     (i      +     gij5     +     g-^g,      +      ^g-,      t       6^.) . 


9 

COMPUTATION. 

2.00000000 

81 
81 
81 

.22222222 
.00274348 
.00003387 
.00000042 

1 
3 
5 

7 

.22222222 
.00091449 
.00000677 
.00000006 

log  4  = 

.22314354 
1.38629436 

.-.  logs  =  1.60943790 

Li  like  manner  we  may  proceed  to  compute  the  logarithms  of  the 
prime  numbers  from  the  formula,  and  obtain  those  of  the  composite 
numbers  on  the  principle  that  the  logarithm  of  the  product  equals 
the  sum  of  the  logarithms  of  the  factors. 

Thus,  the  Napierian  logarithm  of  the  base  of  the  common  system, 
10,  =  log  5  + log  3  =  2.30358508. 


HIGHER    EQUATIONS.  411 

169,  I^rop, — Tlie  modulus  of  the  common  system  is 
.43429448  +  . 

Dem. — Since  the  logarithm  of  a  number,  in  stoy  system,  divided  bv 
the  Napierian  logarithm  of  the  same  number  is  equal  to  the  modulus 
of  that  system  {166),  we  have 

Com.  log  10  ,  ,        , 

— r-^-T7^  =  modulus  of  common  system. 

Nap.  log  10 

But  com.  log  10  =  1,  and  Nap.  log  10  =  2.30258508,  as  found 
above.    Hence, 

Modulus  of  common  system  =  —  =  .43429448 


170.  Prop.— The  NajneriaJi  base  is  2.718281828. 

Dem.— Let  e  represent  the  base  of  the  Napierian  system.     Then 
by  (163) 

com.  log  e  :  Nap.  log  e  : :  .43429448  :  1. 

But  the  logarithm  of  the  base  of  a  system,  taken  in  that  system  is  1, 
since  a^  =  a.  Hence,  Nap.  log  e  =  l,  and  com.  log  e  =  .43429448. 
Now  finding  from  a  table  of  common  logarithms  the  number  corres- 
ponding to  the  logarithm  .43429448,  we  have  e  =  2.718281828. 


SECTION    IV. 

HIGHER     EQUATIONS. 

171,  Since  every  equation  with  one  unknown  quantity, 
and  real  and  rational  coefficients,  can  be  transformed  into 
one  of  the  form 

X-  +  Ax"-^  4-  Bx-~^  +  Cx^-^ L  =  0,  (1) 

this  will  be  taken  as  tlie  typical  numerical  equation  whose 
solution  we  shall  seek  in  this  and  the  succeeding  sections ; 


412  HIGHER    EQUATION-S. 

and  we  shall  frequently  represent  it  by/(^)  =  0,  read 
"function  x  equals  0."  The  notation  /"(a;)  signifies  in 
general,  as  has  been  before  explained  {187),  simply  any 
expression  inyolving  x.  Here  we  use  it  for  this  particular 
form  of  expression.  We  shall  also  use/'  {x)  as  the  symbol 
for  the  first  differential  coefficient  of  this  function. 


172.  JProp, —  When  an  equation  is  reduced  to  the  form 
x"4-Ax"~^  +  Bx"~'^  +  Cx"~^ .  . .  .\i  =  ^,the  roots,  ivith  their 
signs  changed,  are  factors  of  the  absolute  {known)  term,  L. 

[For  demonstration  see  p.  363-] 

Its.  CoE. — If  a  is  a  root  of  f  (s)  =  0,  f  (x)  is  divisible 
by  x—a]  and,  conversely,  ift{x)  is  divisible  by  x—a,  a  is  a 
rootofi{x)^0. 

Dem. — The  first  statement  is  demonstrated  in  tlie  proposition,  and 
tlie  second  is  evident,  since  as  /(;«)  is  divisible  by  x—n,  let  the 
quotient  be  ^(a?);  whence  (,»—«)(* (a-)  -  0.  Now  x=a  will  satisfy 
this  equation,  since  it  renders  x—a=:0,  and  does  not  render  (p{x) 
infinity,  since  by  hypothesis  x  does  not  occur  in  the  denominator.* 


I 


174.  I*rop.—If  the  coefficie7its  and  absolute  term  in 

x"  +  Ax»-'  +  Bx"-'  +  Cx°-^ 1^  =  0,  are  all  integers,  the 

equation  can  have  no  fractional  root. 

Dem. — Suppose  in  this  equation  a?  =  -  ;   .-  being  a  simple  fraction 

*      t 

in  its  lowest  terms.     Substituting  this  value  of  x,  we  have 

i»  ^—1  fn—l  ^n— 3 


*  Could  there  be  a  term  of  the  form  — -  in  6  {x\  x  —  a  would  render  it  oc,  and 

x—a 

(X—a)  <}>  (x)  would  he  0  X  00,  which  is  indeterminate,  since  OxQo  =  Oxi  =  §. 


HIGHER    EQUATIONS.  413 

Multiplying  by  t"^^  we  obtain 

on 

Now,  by  liypotliesis,  all  tlie  terms  except  the  first  are  integral,  and  the 
first  is  a  simple  fraction  in  its  lowest  terms,  as  by  hypothesis  s  and  t 
are  prime  to  each  other.  But  the  sum  of  a  simple  fraction  in  its  low- 
est terms  and  a  series  of  integers  cannot  be  0.     Therefore  x  cannot 

equal  -  ,  a  fraction. 

17i>»   ScH. — This  proposition  does  not  preclude  the  possibility  of 
surd  roots  in  this  form  of  equation.    These  are  possible. 


Its.  Prop.— An  equation  i{x)  =  0  {171)  of  the  nth 
degi^ee,  has  n  roots  {if  it  has  any),  and  no  rnore. 

Dem. — Let  a  be  a  root  of  /  (x)  =  0,  which  is  of  the  nth  degree. 
Dividing /(.t)  hjx  —  a  {17 S),  we  have  (p  {x)  =  0,  an  equation  of  the 
{n  —  l)th  degree. 

Let  6  be  a  root  of  0  {x)  =  0,  and  divide  (p{x)'bj  x  —  b  {173).  Call 
the  quotient  0'  {x),  whence  0'  {x)  =  0,  an  equation  of  the  {n  —  2)th 
degree.  In  this  way  the  degree  of  the  equation  can  be  diminished  by 
division  until,  after  n  —  1  divisions,  there  results  0"  {x)  of  the  first 
degree,  and  the  equation  is  x  —  I  =  0.  Therefore, 
f{x)  =  {x  -  a)  (p{x)  =  (x  —  a){x  —  h)  cp' {x)  =  (x  —  a)  {x  —  b)  (x  —  c) 

(f,"  (,r)  =  {x  —  a){x  —  b){x  —  c) {x  —  I)  =  0  ; 

i.  e.,f{x)  is  resolvable  into  n  factors,  of  the  form  x  —  m. 

Now,  2iS  X  =  a,  ov  X  =  b,  or  X  =  any  one  of  the  quantities  a,  b,  c 
. .  .  .  I,  will  render /(a?)  equal  to  0,  each  one  of  these  will  satisfy  the 
equation /(a;)  =  0.     Therefore  this  equation  has  n  roots. 

Again,  since  it  is  evident  that  we  have  resolved /(.r)  into  its  2^rime 
factors  with  respect  to  x,  there  can  be  no  other  factor  of  the  form 
x  —  m  in/(.i'),  hence  no  other  root  oif{x)  =  0,  and  this  whether  m  is 
equal  to  one  or  more  of  the  roots  a,b,  c  .  .  .  .  n,  or  not.  Therefore 
f{x)  =  0  has  only  n  roots. 

1 77.  Cor.  l.—TIie  polynomial  x°  +  Ax"-'  +  Bx"-"  +  Cx"-^ 

L,  or  f  (x),  =  (x  —  a)  (x  —  b)  (x  —  c)  .  .  .  .  (x  —  1),  in 

which  a,  b,  c  ....  1  are  the  roots  of  i  (x)  =  0. 


414^  HIGHER    EQUATIONS. 

178,  Cor.  2. — The  equation  i{x)  =  0  may  have  2,  3,  or 
even  n  equal  roots,  as  there  is  no  inconsistency  in  su^oposing 
a  =  b,  a  =  b  =  c,  or  a  =  b  =  c  ==  .  .  .  .  1,  i^  the  above 
demonstration. 

179,  Cor.  3. — Imaginary  roots  enter  into  equations 
having  only  real  coefficients,  in  conjugate  pairs ;  that  is,  if 
f  (x)  =  0  has  only  real  coefficients,  if  it  has  one  root  of  the 
form  «  4-  ^V—'^^it  has  another  of  the  form  a  —  §\/  —  1 ; 
or,  if  it  has  one  of  the  form  ^V—h  it  has  another  of  the 
form  —§  V—  1. 

This  is  evident,  since  only  thus  can  /  {x)  =  {x  —  a)  {x  —  b)  {x  —  c) 
....  (a?  —  n);  that  is,  if  one  root,  a  for  example,  is  a  — 13^ — 1> 
there  must  be  another  of  the  form  a  +  (3'\/ —  1,  in  order  that  the  pro- 
duct of  these  two  factors  shall  not  involve  an  imaginary.  Thus, 
[aj_(a  +  /3y'iri]  X  [x  —  (a  —  ^^^^l]  =  x'  —  ^ax  +  (a^  +  /32),a  real 
quantity.  So  also  (a?  — /3y^ —  \){x  +  ii  V — 1)  =  ^^  +  /32,  a  real  quan- 
tity. But  if  the  assumed  imaginary  roots  be  not  in  conjugate  pairs, 
the  product  of  the  factors  {x  —  a)  {x  —  b)  {x  —  c) .  .  .  .{x  —  I)  will  in- 
volve imaginaries. 

180,  Cor.  4. — Hence  an  equation  of  an  odd  degree  must 
have  at  least  one  real  root ;  but  an  equation  of  an  even  degree 
does  not  necessarily  have  any  real  root. 

Cor.  5. — If  an  equation  has  a  pair  of  imaginary  roots, 
the  Tcnoimi  quantities  entering  into  the  equation  may  be  so 
varied  that  the  two  imaginary  roots  shall  first  give  place  to 
tiuo  equal  roots,  and  then  these  to  two  unequal  roots. 

As  shown  above,  imaginary  roots  arise  from  real  quadratic  factors 
in/(^).  Let  x^—2ax  +  b  be  such  a  quadratic  factor,  whence  x'^ — 2ax  + 
b  =  0  satisfies /(;r)  =  0,  and  a  ±  '\/a?  —  h  are  the  corresponding  roots 
otf{x)  —  0.  Now,  if  &  <  a^,  these  roots  are  imaginary.  If,  however, 
b  diminishes  or  a  increases  (or  both  change  thus  together),  when 
b  =.  a?  the  two  imaginary  roots  disappear  and  we  have  in  their  place 


HIGHER    EQUATIOITS.  415 

two  real  roots,  each  a.  If  tlie  same  change  in  a  and  h  continues,  so 
that  a?  becomes  greater  than  6,  the  two  real,  equal  roots  in  turn  give 
place  to  two  real,  unequal  roots.  Now  as  a  and  6  are  functions  of  the 
known  quantities  of  the  equation /(.r)  =  0,  such  changes  are  evidently- 
possible. 

1S1»  By  means  of  the  property  exhibited  in  (187), 
produce  the  equations  whose  roots  are  given  in  the  follow- 
ing examples : 

1.  Roots  1,  —  3,  4. 

2.  Roots  a/2,  —  V2,  —  1,  3. 

3.  Roots  1,  2,  2,  —  3,  4. 

4.  Roots  —  3,  2  +  V^^l,  2  —  V^^' 
6.  Roots  3,  —  2,  —  2,  —  2,  1. 

6.  Roots  f ,  i  -  |. 

7.  Roots  1  ±  V^^,  2  ±  V^^. 

8.  Roots  1J-,  2,  V3,  —  V^- 

9.  Roots  V—  2,  —  V—  2,  VS,  —  a/5. 

10.  Roots  10,  —  13,  I,  1. 

11.  Roots  3  —  2  a/3,  3  +  2^/3,  2  —  3  a/^^,  2  +  3 

V~^l,  1,  -  1. 


182.  Prop, — If  the  equation  f  (x)  =0  has  equal  roots, 
tlie  Mgliest  common  divisor  o/f  (x)  a}id  its  differential  coeffi- 
cient,"^ f  (x),  being  ptU  equal  to  0,  constitutes  an  equation 
luMch  has  for  its  roots  these  equal  roots,  and  mo  other  roots.\ 

*  The  differential  coefficient  of  a  function  is  sometimes  called  its  first  derived 
polynomial, 

The  student  must  not  suppose  that  the  roots  off  (or)  =  0,  and  its  first  diflJeren- 
tial  coefficient/' (a?)  =  0,  are  necessarily  alike.  f'{x)  —  a  series  of  terms  some  of 
which  may  be  +  and  some  — ,  and  which  may  destroy  each  other,  so  as  to  render 
fix)  3=  0,  for  other  values  of  x  than  such  as  render/ (a:)  —  0,  and  not  necessarily 
for  any  which  <io  render/  {x)  =  0,  except  the  equal  roots  of  the  latter. 


416  HIGHEE  EQUATIONS. 

Dem. — Let  a  be  one  of  tlie  m  equal  roots  of  fix)  —  0,  and  let  the 
other  roots  be  h,  e  .  .  .  .  I;  then  /(.i-)  =  {x  —  a)"^{x  —  h)  (x  —  c)  .  . .  . 
{x  —  I)  {177)-    Differentiating  {14:0)  and  dividing  by  dx,  we  have 

/'  {x)  =:m{x  —  ay^-^  (x  -  b)  (x  -  c) .  .  .  .  (:x ~  l)  +  {x  -  a)m  {x  —  c) . , ,  , 
{x  —  I)  +  ....  +  {x  —  a)'"-  {x  —  b)  {x  —  c)  .  . .  .  +  etc. 

Now  (x  —  «)"*~^  is  evidently  the  highest  common  divisor  of  f{x)  and 
f  (x),  and  (x  —  a)^~'^  =  0  is  an  equation  having  a  for  its  root,  and 
having  no  other. 

In  a  similar  manner,  if  /(^)  =  0  has  two  sets  of  equal  roots,  so 
that 

f{x)  =  (x  —  a)'n  {x  —  by{x  —  c){x  —  d) (x  —  I), 

differentiating  and  dividing  by  dx,  we  have 

/■  {x)  =  m{x  —  a)"^-^  {x  —  hy{;x  —  c){x  —  d) (x  —  l)  +  {x  —  a)"" 

r{x  —  by-^  {x  —  c){x  —  d) . . .  .{x  —  T) 

+  {x  —  a)m  {x  —  by{x  —  d)  .  .  .  .{x  —  n)  -\-  {x  —  a)m  (x  —  by  (x  —  c) 
. .  .  .  {x  —  l)  + . .  .  .  +  {x  —  a)"^  (x  —  by{x  —  c){x  —  d)..  .  .  +  etc. 

Now  the  highest  common  divisor  of  f{x)  and  f\x)  is  evidently 
{x  —  ft)"*-'  (x  —  &)'"-^  Putting  this  equal  to  0,  we  have  {x  —  a)"'-'^ 
{x  —  by-'^  =  0,  an  equation  which  is  satisfied  hjx  =  a  and  x  =  b,  and 
by  no  other  values. 

Thus  we  may  proceed  in  the  case  of  any  number  of  sets  of  equal 
roots. 

183,  Sen. — In  searching  for  the  equal  roots  of  equations  of  high 
degree,  it  may  be  convenient  to  apply  the  process  of  the  proposition 
several  times.  Thus,  suppose  that  f{x)  =  0  has  7n  roots  each  equal 
to  a,  and  r  roots  each  equal  to  b.  Then  the  highest  common  divisor 
of /(cc)  and /'(a?)  is  of  the  form  {x  —  a)"*-^  {x  —  6)'—^ ;  whence  {x — a)""-^ 
(x  —  by-'^  =  0  is  an  equation  having  the  equal  roots  sought.  There- 
fore we  can  find  the  highest  common  divisor  of  (x — a)'^~'^(x — &)'■~^  and 
its  differential  coefficient  which  will  be  of  the  form  (x — a)"'-%x—by-^, 
and  write  {x  —  a)"^^  (a,  —  j^r-s  _  q^  as  an  equation  containing  the 
roots  sought.  This  process  continued  will  cause  one  of  the  factors 
{x  —  a)  or  {x  —  b)  to  disappear  and  leave  (x  —  «)'"—'"  —  0,  when  m  >  r; 
{x  —  &)'•-'"  =  0,  when  r  >  m  ;  or  (x  —  a)  (x  —  b)  =  0,  when  m  =  r. 
From  any  one  of  these  forms  we  can  readily  determine  a  root. 


HIGHER    EQUATIOJ^S.  417 

184,  ^vop, — In  an  equation  f  (x)  =0,  f(x)  ivill 
change  sign  luhen  x  passes  through  any  real  root,  if  there  is 
iut  one  such  root,  or  if  there  is  an  odd  number  of  such 
roots ;  lut  if  there  is  an  eveis"  numler  of  such  roots,  f  (x) 
will  not  change  sign. 

Let  a,  1),  c  ....  e  be  tlie  roots  of  f{x)  =  0,  so  that  f((c)  =  {x—a) 
(x—h)  {x—c)  ....  (x—e)  =  0  {It 7)'  Ck)iiceive  x  to  start  with  some 
value  less  tlian  tlie  least  root,  and  continuously  increase  till  it 
becomes  greater  tlian  tlie  greatest  root.  As  long  as  x  is  less  tlian  tbe 
least  root,  all  the  factors  x—a,  x—b,  etc.,  are  negative;  but  when  x 
passes  the  value  of  the  least  root,  the  sign  of  the  factor  containing 
that  root  will  become  +,*  and  if  there  is  no  other  equal  root,  this 
factor  will  be  the  only  one  which  will  change  sign.  Hence  the  pro- 
duct of  the  factors  will  change  sign.  But  if  there  is  an  even  number 
of  roots,  each  equal  to  this,  an  even  number  of  factors  will  change 
sign ;  whence  there  will  be  no  change  in  the  sign  of  the  function. 
If,  however,  there  is  an  odd  number  of  equal  roots,  the  passage 
of  X  through  the  value  of  this  root  will  cause  a  change  of  sign 
in  an  odd  number  of  factors,  and  hence  will  change  the  sign  of  the 
function. 

Finally,  as  it  is  evident  that  the  signs  of  the  factors,  and  hence  of 
the  function,  will  remain  the  same  while  x  passes  from  one  root  to 
another,  and  in  all  cases  changes  or  does  not  change  as  above  when  x 
passes  through  a  root,  the  proposition  is  established. 

The  following  example  will  be  found  very  instructive  :  x^  +  4x*— 
14t;"-'— 17a;— 6  =  0.  The  least  root  of  this  equation  is  — 3.  When  x< 
—d,f{x)  is  — ;  when  x  =  —S,f{x)  =  0;  when  x  passes  —3,  increas- 
ing,/(a*)  changes  from  —  to  +,  and  remains  +  till  x  =  —1,  when  it 
becomes  0,  and  changes  sign  as  x  passes  —1,  iioticitJiStanding  they  are 
equal  roots.  But  there  is  an  odd  number  of  such  roots,  viz.,  three. 
But  in  a^— 14c2  +  64a;— 96  =  0,  two  equal  roots  of  which  are  4,  if  we 
substitute  2  we  get  f{x)  =  —16,  and  substituting  5,  f{x)  =  —  1,  the 
function  not  changing  sign,  although  a  root  has  been  passed. 


*  Suppose  c  be  the  least  root,  and  that  c'  is  the  next  state  of  x  greater  than  c ; 
then  </—€  is  + . 


418  HIGHEE    EQUATION'S. 

185,  I^rop» — Changing  the  signs  of  the  terms  of  an 
equation  containing  the  odd  powers  of  the  imknoivn  quan- 
tity changes  the  signs  of  the  roots. 

Dem. — If  x—a  satisfies  tlie  equation  x^—Axl^  +  B^x^—Cx  +  D  =  0, 
we  liave  a^—Aa^-\-Ba^—Ca  +  D  =  0.  Now  changing  the  signs  of  the 
terms  containing  the  odd  powers  of  a?,  we  have  x^—Aa^—Bx^+Cx  + 
Z>  =  0.  This  is  satisfied  hy  x  =  —a,ii  the  former  is  by  a?  =  a.  For, 
substituting  —a  for  x,  we  have  a^—Aa'^  +  Ba^—Ca  +  D  =  0,  the  same 
as  in  the  first  instance. 

186.  Cor. — Changing  the  signs  of  the  terms  containing 
the  even  potvers  will  ansiver  equally  ivell,  since  it  amounts 
to  the  same  thing  ;  and  if  we  are  careful  to  put  the  equation 
i7i  the  complete  form,  changing  the  signs  of  the  alternate 
terms  ivill  accomplish  the  purpose. 

III. — The  negative  roots  oi  x^—lx  +  Q  =0,  are  the  positive  roots  of 
— a;^  +  7:c  +  6  =  0,  or  of  x^—lx—Q  =  0  (0  being  considered  an  even 
exponent)  ;  or,  writing  the  equation  x^±{)x''-—lx  +  Q  —  0,  changing  the 
signs  of  alternate  terms,  and  then  dropping  the  term  with  its  coeffi- 
cient 0,  we  obtain  the  same  result. 


in,  the  negative  roots  of  x^  —  *lx^  —^x^->rQx^—\^2x-  +  ^0?>x-- 
240  =  0,  are  the  positive  roots  of  x^  +  1lx^-^x^-%x^-\Z2x:^—b^^x-2^<(i 
=  0,  or  of  —x^-7x^  +  5x^  +  8x^  +  132.c-  +  508^  +  240  =  0. 


187,  Frob. — To  evaluate"^  f  (x)  for  any  particular 
value  of  X,  as  X  ^  a,  more  expeditiously  than  hy  direct  sub- 
stitution. 

Solution.— As /(:?•)  is  of  the  form  x'^  +  Ax''^^  +  Bx""-"^  +  Cx''-^ .  .  L, 
let  it  be  required  to  evaluate  x'^  +  Ax^  +  Bx^  +  Cx  +  I>  for  x  =  a.  Write 
the  detached  coefficients  as  below,  with  a  at  the  right  in  the  fonn  of 
a  divisor ;  thus 

*  This  means  to  find  the  value  of.  Thus,  suppose  we  want  to  find  the  vahie  of 
a?*— 5cK°  +  2cc*— 3aj*  +6a;^— a;~12,  for  x  =  5.  We  might  substitute  5  for  x,  of  course, 
and  accomplish  the  end.  But  there  is  a  more  expeditious  way,  as  the  solution  of 
this  problem  will  show. 


HIGHER    EQUATIONS.  419 

1     +A  +B  +G  +D  \a__ 

a  a^  +  Aa  a^  +  Aa^  +  Ba  a^  +  Aa^+Ba'^  +  Ca 


a  +  A     a'  +  Aa  +  B     a^  +  Aa-  +  Ba  +  C     a'^  +  Aa^  +  Ba- +  Ca  +  B 

Having  written  tlie  detaclied  coefficients,  and  the  quantity  a  for  which 
f{.i')  is  to  be  evaluated  as  directed,  multiply  the  first  coefficient  1  hy 
a,  write  the  result  under  the  second,  and  add,  giving  a  +  A.  Multiply 
this  sum  by  a,  write  the  product  under  the  third  coefficient  B,  and 
add,  giving  a^  +  Aa  +  B.  In  like  manner  continue  till  all  the  coeffi- 
cients (including  the  absolute  term,  which  is  the  coefficient  of  x^) 
have  been  used,  and  we  obtain  a*  +  Aa^  +  Ba^  +Ca  +  B,  which  is  the 
value  otf{x)  for  x  =  a. 

Illustration.— To  evaluate  x^  —5x^  +  2x^  —  dx^  +  6^  —a;— 12,  for 


X  =  iy: 

1 

-5 

+  2 

-3 

+  6 

-1 

-12  15 

'5 

0 

10 

35 

205 

1020 

0 

2 

7 

41 

204 

1008 

Now  1008  is  the  value  of  x^—5x^  +  2x^—Sx^  +  ex^—x—12,foTX  =  5; 
and  it  is  easy  to  see  that  much  labor  is  saved  by  this  process. 

We  are  now  prepared  for  the  solution  of  the  following 
important  practical  problem : 

188.  ^Proh, — To  find  the  commensuraUe  roots  of  numer- 
ical higher  equations. 

The   solution  of    this    problem   we   will    illustrate  by    practical 
examples. 


EXAM  PLES. 


1.  Find  the  commensurable  roots  of  x^—2o!:^^16x^-{-^x^-{- 
68.T  +  48  =  0,  if  it  has  any. 

Solution. — By  (17 d),  if  this  equation  has  any  commensurable 
roots  they  are  integral ; — it  can  have  no  fractional  roots. 

Again,  by  {172),  the  roots  of  this  equation  with  their  signs 
changed  are  factors  of  48.  Now,  the  integral  factors  of  48  are  1,  2,  3, 
4,  6,  8,  12,  16,  24,  48.    Hence,  if  the  equation  has  commensurable 


420  HIGHER    EQUATIONS. 

roots,  tliey  are  some  of  these  numbers,  with  either  the  +  or  —  sign. 
We  will,  therefore,  proceed  to  evaluate  f{x){i.  e.,  in  this  case  x^ — 
2«4-15«3  +  8a;-^  +  68a;  +  48),  f  or  a?  =  +1,  x  = -l,x  =  +2,  x  =  -2,  etc., 
by  (i^ 7),  as  follows: 

1    -2  -15  +  8  +68  +48  I  +1 

_i  ni         Zli?  IL^  _J2 

-1  -16         -  8  60  108 

Hence  we  see  that  f or  cc  =  + 1,  f{x)  =  108,  and  + 1  is  not  a  root  of 
f[x)  =  0.     Trying  a;  =  —  1,  we  have 


-2 

-15 

+  8 

+  68 

+  48 

-1 

3 

13 

-20 

-48 

-3 

-13 

20 

48 

0 

-15 

+  8 

+  68 

+  48  1 

+  3 

0 

-30 

-44 

+  48 

-15 

-23 

24 

96 

Thus  we  see  that  for  x  =  —l,f(x)  =  0,  and  hence  that  —1  is  a  root 
of  our  equation. 

We  might  now  divide /(.r)  by  x  +  1  (173)  and  reduce  the  degree 
of  the  equation  by  unity.  But  it  will  be  more  expeditious  to  proceed 
with  our  trial.     Let  us  therefore  evaluate /(a^)  for  x  =  +2.    Thus  : 

1     -3 

_2 

0 

Hence  for  x  =  +2,f{x)  =  96,  and  +3  is  not  a  root.     Trying  a;  —  —3, 
we  have 

1     -3  -15  +  8  +68  +48  |_-3 

-3  _8  _14  -44  -48 

-4-7  23  24  0 

Hence  for  x  =  —2,  f{x)  =  0,  and  —3  is  a  root.      Trying  x  =  +3, 
we  have 

1     -3  -15  +  8  +68  +48  |_+3 

_3  _3  -36  -84  -48 

1  -13  -28  -16  0 

*  Of  course  it  is  not  necessary  to  retain  the  +  sign,  as  we  have  done  in  the  pre- 
ceding operations :  it  has  been  done  simply  for  emphasis. 


HIGHER    EQUATIONS.  421. 

Hence  for  x  =  +S,  f(x)  —  0,  and  +3  is  a  root.      Trying  x  =  —3, 
we  have 

1     -3  -15  +8  +68  +  48  |_--3 

-3  15  0  -24  -133 

-5  0  8  44-84 

Hence  for  x  =  —S,f(x)  =  —84,  and  —3  is  not  a  root.     Trying  a;  =  4, 
we  have 

1     -3  -15  +  8  +68  +48  1^ 

4  _8  -28  -80  -48 

3  -  7  -30  -13  0 

Hence  for  x  =  4,/(x')  =  0,  and  4  is  a  root. 

We  have  now  found  four  of  the  roots,  viz.,  —1,  —3,  3,  and  4. 
Their  product  with  their  signs  changed  is  24.  Hence,  by  {172)  48-f- 
24  =  3  is  the  other  root  with  its  sign  changed,  i.  e.,  there  are  tico 
roots  —3. 

That  our  equation  had  equal  roots  could  have  been  ascertained  by 
the  principle  in  {182) ;  but  as  the  process  of  finding  the  H.  C.  D.  is 
tedious,  it  is  generally  best  to  avoid  it  in  practice. 

to  11.  Find  the  roots  of  the  following : 

2.  a^—a^—39x^-{-24:X -{-180  =  0; 

3.  x^-{-5x^—9x—4:6  =  0; 

4.  a^-\-2x^—2dx—60  =  0; 

5.  a^—dx^—Ux^  +  ^8x—32  =  0; 

6.  x^—Sx^-\-ldx—6  =  0; 

7.  cf^—llx^-\-lSx—S  =  0;^ 

8.  x^—dx^-{-6a^—3x^—dx  +  2  =  0; 

9.  a^—lSx^-\-6W—171x^-i-216x— 108  =  0; 

10.  a^—4:6x^—4.0x-\-84:  =  0; 

11.  a^—3x^—da^  +  21x^—10x  +  2i  =  0. 


*  In  order  to  apply  the  process  of  evaluation,  the  coefficients  of  the  missing 
powers  must  be  supplied.    Thus  we  have  1  +  0—11  + 18—8. 


422  HIGHER    EQUATIONS. 

12  to  17.  Apply  the  process  for  finding  equal  roots  (192, 
183)  to  the  following: 

12.  a;3  +  8a;2  +  20a;  +  16  =  0; 

13.  x^—x^—^x  +  VZ  =  0; 

14.  a;3_5^2_8ic  +  48=:0; 

15.  a;4_ii^2_|.i8^_8  z=0; 

16.  ic4_j_i3^3_f_33^2_j_31a;4.io=:  0; 

17.  aP—ldx^  +  QW—l'ilx^+^lQx—1^^  =  0. 

18  to  24.  Having  found  all  but  two  of  the  roots  of  each 
of  the  following  by  (187),  reduce  the  equation  to  a  quad- 
ratic by  (Its),  and  from  this  quadratic  Qnd  the  remaining 
roots : 

18.  x^—Qx^  +  lOx—%  =  0; 

19.  a;4_4a:3_8a;-}-32  =  0; 

20.  a^—Sx^-\-x-{-2  =  0; 

21.  x^—Qx^  +  Ux— !&  =  ()', 

22.  Q^—l^T^^bW—Ux  +  A:^  =  0  ;* 

23.  a;4_9^  +  i7a;2_^27a:— 60  =  0; 

24.  0:5— 42;^— 16:^3  _|_ii2a;2— 2082;  + 128  —  O; 


25.  2x^—dxi-\-2x—^  =  0;  t 

26.  3a:3— 22;2— 62;  +  4  =  0; 

27.  8^:3— 262;2+lla;  +  10  =  0; 

28.  x^—^x-{-^  =  0 ;     (Look  out  for  equal  roots.) 

29.  c^—6x^-\-9ix^—dx-]-4:i^  =  0. 


*  Apply  the  method  for  finding  equal  roots. 

+  We  have  a;'— '^je'  +  a;— -  =  0.    Put  »  =  |i  '^^^"^^  |^ ~ 2^ ^'^  '^k^~i  ~^' 

or.  or,  3 

or  y^—~-^y^  +  Vy—-Y  =  0-    If  now  ^  =  2,  we  have  y^—Sy^'  +  Ay—n  =  0,  which 


can  be  solved  as  before,  for  one  value  of  y,  and  the  equation  then  reduced  to  a  quad- 
ratic and  solved  for  the  ( 
the  values  of  x  required. 


ratic  and  solved  for  the  other  values.    Finally,  remembering  that  x  =  ^y,  we  have 


INTERPRETATION    OE    EQUATIONS.  4^ 

SECTION    V. 

DISCUSSION,  OR  INTERPRETATION,  OF  EQUATIONS. 

ISO.  To  jyiscusSf  or  Interpret,  an  Equation 
or  an  Algebraic  Expression,  is  to  determine  its 
significance  for  the  various  values,  absolute  or  relative,  wbich 
may  be  attributed  to  the  quantities  entering  into  it,  with 
special  reference  to  noting  any  changes  of  value  which  give 
changes  in  the  general  significance. 

Such  discussions  may  be  divided  into  two  classes:  1st. 
The  discussion  of  equations  or  expressions  with  reference 
to  their  constants ;  and  2d.  The  discussion  of  equations  or 
expressions  with  reference  to  their  variables. 

The  following  principles  are  of  constant  use  in  such  dis- 
cussions : 

190.  Prop. — A  fraction,  when  compared  with  a  finite 

quantity,  becomes : 

1.  Equal  to  0,  tvhen  its  numerator  is  0  and  its  denomina- 
tor finite,  and  ivhen  its  numerator  is  finite  and  its  denomi- 
nator 00  . 

2.  Equal  to  oo ,  luhen  its  numerator  is  finite  and  its  de- 
nominator 0,  and  when  its  numerator  is  oo  and  its  denomi- 
nator finite, 

3.  It  assumes  an  indeterminate  form  when  numerator  and 
denominator  are  loth  0,  and  when  they  are  both  oo  .* 

Dem. — These  facts  appear  when  we  consider  that  the  value  of  a 
fraction  depends  upon  the  relative  magnitudes  of  numerator  and  de- 
nominator. 


*  By  this  it  is  meant  that  -  and  —  may  have  a  variety  of  values,  not  that  they 
necessarily  do  have. 


424  IKTERPEETATIOK    OF    EQUATIOKS. 

1.  Let  a  be  any  constant  and  x  a  variable,  then  tlie  fraction  - 

''  a 

diminishes  as  x  diminishes,  and  becomes  0  when  x  is  0.  Again,  the 
fraction  ~  diminishes  as  x  increases,  and  when  x  becomes  oo,  i.  e., 

a 
greater  than  any  assignable  magnitude,   -   becomes  less  than  any 

assignable  magnitude  or  infinitesimal,  and  is  to  be  regarded  as  0  in 
comparison  with  finite  quantities.  (See  139  and  14:8,  Dem.,  and 
foot-note.) 

x 

2.  As  X  increases,  the  fraction  -  increases,  and  hence  when  x  be- 

CL 

comes  infinite,  the  value  of  the  fraction  is  infinite.    Also,  as  x  dimin- 

a 
ishes,  the  value  of  -  increases;  hence,  when  x  becomes  infinitely 

small,  or  0,  the  value  of  the  fraction  exceeds  any  assignable  limits, 
and  is  therefore  oo . 

X 

3.  Finally,  if  x  and  y  are  variables,  -  diminishes  as  x  diminishes, 

and  increases  as  y  diminishes.    What  then  does  it  become  when  x=0, 

0 

and  2/  =  0?  i.  e.,  what  is  the  value  of  x?     Simple  arithmetic  would 

0 

lead  us  to  suppose  that  t,  was  absolutely  indeterminate,  i.  e. ,  that  it 

0 
might  have  any  value  whatever  assigned  to  it,  for    n  =  5,     since 

0 

0  =  5x0  =  0;  Q  =  ''',  since  0  =  7x0  =  0,  etc.    But  a  closer  inspection 

0 

will  enable  us  to  see  that  the  symboi  x  is  not  necessarily  indetermi- 
nate, or  rather  that  the  expression  which  takes  this  form  for  parti cu-' 
lar  values  of  its  components,  has  not  necessarily  an  indefinite  number 
of  values  for  these  values  of  its  components.     Thus,  what  the  value 

X 

of  -  will  be  when  x  and  y  each  diminish  to  0  will  evidently  depend 
y 

upon  the  relative  values  of  x  and  y  at  first,  and  which  diminishes  the 

X 

faster.     Suppose,  for  example,  that    y  =  ^x\    then       =  ^ .    Now, 

y      ox 

suppose  X  to  diminish ;  the  denominator  will  diminish  5  times  as  fast 
as  the  numerator,  and  whatever  the  value  of  x,  the  value  of  the  frac* 


tfc^TEEPIlETATION"    OF    EQUATIONS.  425 


tion  will  be  \.     So  if   y  =  7x,   -  =  =- ,  wMch  is  |  for  any  value  of  x. 

if       •'*' 

X      0       X       ^ 
Hence,   when    x  =  0,     and    v  =  0,     we  liave    -  =  ^  =  --  =  -  ^    or 
■^         '  y      {)      ox      0 

X      0       X       ^  x      0 

-  =  Tr  =  ^^  =  -  ^  or  -  =  7,  =  any  other  value  depending  upon  the  rel- 

X      x> 
ative  values  of  x  and  ?/.     So,  also,  if  a;  =  oo ,  and  2/  =  oo ,  -  —  —  ;  but 

J  >  if  '    y  (X)   ' 

a?      00        a;       1 
if  y  =  6x,  we  have  -  =  —  z=  ^r^  =  ^.     And  so  if  y  —  lOa",  we  have 

a;       00  x  1 

=  —  =  ZTYT  =  T7i  •  Thus  we  see  that  the  mere  fact  that  numerator 
y      00       lOa*      10 

and  denominator  become  0,  or  become  oo ,  does  not  determine  the  value 
of  the  fraction,  i.  e.,  gives  it  an  indeterminate  form. 

191,  A  Ileal  J^iiniber  or*  Quantity  is  one  which 
may  be  conceived  as  lying  somewhere  in  the  series  of  num- 
bers or  quantities  between  —  oo  and  +  oo  inclusive. 

III. — Thus,  if  we  conceive  a  series  of  numbers  varying  both  ways 
from  0,  i.  e.,  positively  and  negatively  to  oc ,  we  have 

-  00  ....  -4,  -3,  -2,  -1,  0,   +1,   +2,  +3,  +4,  .  .  .  .   +  oo. 

Now  a  real  number  is  one  which  may  be  conceived  as  situated 
somewhere  within  these  limits  ;  it  may  be  +  ,  — ,  integral,  fractional, 
commensurable,  or  incommensurable.  Thus,  +15624  and  —15624 
will  evidently  be  found  in  this  series.  +^^  may  be  conceived  as 
somewhere  between  +  5  and  +  6,  though  its  exact  locality  could  not  be 
fixed  by  the  arithmetical  conception  of  discontinuous  number.  So, 
also,  —y- is  somewhere  between  —5  and  —6.  Again,  +  ^^5  is  some- 
where between  +  2  and  +  3,  though,  as  above,  we  cannot  locate  it 
exactly  by  the  arithmetical  conception. 

The  following  Geometrical  Illustration  is  more  complete  than  the 
arithmetical.  Thus,  let  two  indefinite  lines,  as  CD  and  AB,  intersect 
(cross)  each  other,  as  at  0.  Now  let  parallel,  equidistant  lines  be 
drawn  between  them.  Call  the  one  at  <z  +1,  that  at  b  will  be  +  2,  at 
c  +3,  etc.  So,  also,  the  line  at  a'  being  —1,  that  at  V  will  be  —2,  at 
d  —3,  etc.  Now  conceive  one  of  these  lines  to  start  from  an  infinite 
distance  at  the  left  and  move  toward  the  right.     When  at  an  infinite 


426 


II^TERPRETATIOK    OF    EQUATIONS. 


distance  to  the  left  of  0  its  value  would  be  —  oo ,  and  in  passing  to  0 
it  would  pass  through  all  possible  negative  values.  In  passing  O  it 
becomes  0  at  0,  changes  sign  to  +  as  it  passes,  and  moving  on  to 


^ 

r^ 

b-'' 

1^^ 

--^« 

^ 

B    - 

7     - 

6     - 

5    - 

4-3 

-  2 

2      +3     +4-1-5     +0      +7      + 

8     + 

8+9- 


infinity  to  the  right,  passes  through  all  possible  positive  values.  Hence 
we  see  how  all  real  values  are  embraced  between  —  cx>  and  +  oo ,  in- 
clusive.* 

192,  An  Imaginary  JV^uniber  or  Quantity  is 

one  which  cannot  be  conceived  as  lying  anywhere  between 
the  limits  of  —  oo  and  +  oo ,  as  explained  above.  The 
algebraic  form  of  such  a  quantity  is  an  expression  involving 
an  even  root  of  a  negative  quantity.f 


EXAM  PLES. 


1.  "What  are  the  values  of  x  and  y  in  the  expressions 

1)'  —  1)             aV  —  a'l)       ,        7       ,,       T  ,    , 

X  = , ,    if  = -r- ,  when  o  =  o  and  a  and  a  are 


a  —  a 


a 


a; 


*  For  example,  the  student  who  is  acquainted  with  the  elements  of  geometry 
knows  how  to  construct  a  line  which  is  exactly  equal  to  V^  (Geom.,  Part  1, 11<¥). 
This  line  he  can  locate  between  +2  and  +-3,  and  also  between —2  and —3,  since 
|/5  is  both  +  and  — . 

+  Transcendental  functions  afford  other  forms  of  imaginary  expressions ;  for 
example,  sin-^  2,  sec^  >^,  log  (—120),  log  (— w),  etc.  But  our  limits  forbid  the  con- 
sideration of  the  interpretation  of  imaginaries,  except  in  the  most  restricted  sense, 
as  indicating  incompatibility  with  the  arithmetical  sense  of  the  problem. 


IKTEEPRETATIOI^    OF    EQUATI0:N'S.  427 

unequal  ?  When  h  =^b'  and  a  =^  a'  ?  When  a=z  a'  and 
h  and  b'  are  unequal  ?  What  are  the  signs  of  x  and  y  when 
by  h'  and  a  >  «',  the  essential  signs  of  a,  a' ,  b,  and  b' 
being  +  ?  When  b>  b'  and  a  <ia"t  If  a'  and  ^  are 
essentially  negative,  and  a  =  a',  and  b  =  &',  what  are  the 
values  of  x  and  y?    If  «'  and  b'  are  each  0  ? 

2.  What   general    relation   between    a   and    a'  renders 

-,  =  0  ?    What  renders  it  oo  ? 


1  -\-  aa 

Solution. — To  render ,  =  0,  we  must  have  a' —a  =  0,  and 

\-\-aa'        ' 

1  +  aa'  finite  or  infinite ;  or  else  we  must  have  l+ad  =  <x) ,  while 
a'— a  is  finite  or  0  {li)0).     Now  a'—a  =  Q  gives  a'  =a',   whence, 

= ; ,  which  is  0  for  any  value  of  a,  finite  or  infinite. 

Hence  the  relation  a'  =  a  fulfills  the  first  requirement.  Let  us  now 
see  if  l  +  aa'  =  oo  will  also  fulfill  this  requirement.  This  gives 
aaJ  =  00 ,  since  subtracting  1  from  oo  would  not  make  it  other  than  oo . 

Thus  we  have  a'  =  —  .    Hence,  for  all  finite  values  of  a  (including  0) 

a'  is  00 ,  and  :; :  =  — ;    =  - :  which  can  only  be  0  when  a  =  co. 

1  +  aa'      aa'        a  '' 

Therefore  the  varticular  values  a'  =  co  =  a  =  cc ,  render :  =  0 ; 

^  1  +  aa' 

but  no  general  values  do,  unless  a  =  a'. 

Aofain,  in  order  that  ~ r  =  oo  ,  we  must  have  1+aa'  =  0,  and 

°  1+aa' 

a'— a  finite  or  infinite;  or  else  we  must  have  a' — a  =  oo,  and  1  +  aa' 
finite  or  0.     Now  1  +  aa'  =  0  gives 

«'  +  ^ 
__L.     <^— ^  _        a'  _a"  +  l  _a"'  +  l_ 
~      a'  *    1  +  aa'  ~~        a'  ~  a' — a'  ~~     0      ~     ' 

a' 

*  This  reduction  is  made  by  dropping  a  and  1,  since  the  subtraction  of  a  finite 
from  an  infinite,  or  the  addition  of  a  finite  to  an  infinite,  does  not  change  the  char- 
acter of  the  infinite.  Thus,  in  this  case,  to  assume  that  dropping  a  and  1  afl"ected 
the  relation  between  numerator  and  denominator,  would  be  to  assign  to  a  and  1 
some  values  with  respect  to  the  infinite  a'.  But  this  is  contrary  to  the  definition  of 
an  infinite. 


428  INTERPRETATION    OF    EQUATIONS. 

for  any  value  of  oJ ,  finite  or  infinite.     Therefore  the  general  relation 

«  = ,  between  a  and  o!  renders  i :  =  go  .*    Let  us  now  see  if 

a  1  +  ad 

the  relation  a' — a  —  oo  will  do  the  s^^me.    Now  if  d — a  =  go  ,  one  or 

the  other  {aJ   or  d)  must    be    go  .     Let    d  =  co.    We    then    have 

;  =  — ;  =:  - ,  which  can  only  be  oo  when  a  =  0.    Hence  the  par- 

1  +  ad      ad      a 

ticular  values  d  =  co  and  a  =  Q  render  zr— — ,  =  oo ,  but  no  general 

1  +  aa'  '  ^ 

values  meet  the  requirement  unless  a  = j . 

3.  What  general    relation    between    a    and    a'    renders 
1  —  aa' 


a!  ■\-  a 


=  0  ?    What  renders  it  oo  ? 


4.  In  the  expression  y  ^z  _  2^  +  4  ±  \x^  —  4^  —  5, 
how  many  values  has  y,  in  general,  for  any  particular  value 
of  :j  ?  For  what  value  or  values  of  cc  has  ?/  but  one  value  ? 
For  what  values  Qix\%y  real  ?  For  what  imaginary  ?  For 
what  values  of  a;  is  ^  positive  ?    For  what  negative  ? 

Solution. — Writing  the  expression  thus, 

we  see  that  the  value  of  y  is  made  up  of  two  parts,  viz.,  a  rational 
part  —  (2a; — 4),  and  a  radical  part  '\/x'' — ix — 5.  But  the  radical  part 
may  be  taken  with  either  the  +  or  the  —  sign.  Hence,  in  general, 
for  any  particular  value  of  x  there  are  two  values  of  y.  3d.  But  if  such 
a  value  is  given  to  x  as  to  render  the  radical  part  0,  for  this  value  of  x, 
y  will  have  but  one  value,  viz. ,  the  rational  part.  But  the  condition 
^x'^ — ix — 5  —  0  gives  a;  =  5  and  — 1.  Thus,  for  a;  =  5,  y  —  —6,  but 
one  value;  and  for  x  —  — 1,  2^=  +6,  also  but  one  value.  3d.  To 
ascertain  for  what  values  of  x,  y  is  real,  we  observe  that  y  is  real  when 
x^ — 4a; — 5  is  positive,  and  imaginary  when  a;^— 4a; — 5  is  negative. 
Now  for  X  positive,  x^ — (4a; +  5)  is  +  when  a;^  >4a;  +  5 ;  and  for  x  nega- 

*  It  is  to  be  observed  that  the  relation  a  —  — -,  requires  that  a  and  a'  have  dif- 
ferent essential  signs  ;  while  the  relation  a'  —  a  requires  that  they  have  the  same 
essential  signs. 


IKTERPKETATIOK    OF    EQUATIOiq-S.  429 

tive,  we  have  x'^  +  4x—5,  which  is  positive  when  x-  +  Ax>5.  The 
former  inequality  gives  x^ — ix  +  4 > 9,  or  x>5;  and  the  latter  gives 
x^  +  4x  +  4:>9,  or  a;  >  1.  Hence  for  positive  values  of  x  greater  than  5, 
y  is  real,  and  for  negative  values  of  x  numerically  greater  than  1,  y  is 
real.  The  4th  inquiry  is  answered  by  this  :  y  is  imaginary  for  all 
values  of  a;  between — 1  and  +5.     5th.  To  ascertain  what  +  values  of 


X  render  y  + ,  and  what  — ,  we  observe  that  —  {2x — 4)  ±  -y/a;^ — ix — 5 
can  only  be  +  when  the  +  sign  of  the  radical  part  is  taken  and  when 
y^x^ — 4a?— 5  >  2a; — 4.  This  gives  x<2  ±  ^/—B,  i.e.,  an  imaginary 
quantity.  Hence  y  is  never  +  for  a;  + .  Taking  the  negative  sign  of 
the  radical,  we  see  that  both  parts  of  the  value  of  y  are  — ,  and  conse- 
quently y  is  real  and  negative  for  all  +  values  of  x  which  render  y 
real,  i.  e.,  for  values  greater  than  5.  Finally,  for  x  —  we  have 
^  =  2a;4-4  ±  's/x^^^ — 5.  Now  when  we  take  the  +  sign  of  the  radi- 
cal, both  parts  are  +  ;  hence  this  value  of  y  is  always  plus.  When 
we  take  the  —  sign  of  the  radical,  y  is  negative  if  2a;  +  4  <  ^/x'^  +  4x — 5. 
But  this  gives  a;<  —  2  ±  /y/ — 3.  Hence  y  is  never  negative  for  any 
negative  value  of  x.  Therefore  both  values  of  y  are  positive  and  real 
for  all  negative  values  of  x  numerically  greater  than  1. 

5  to  22.  Discuss  as  above  the  values  of  y  in  the  follow- 
ing ;  i.  e.,  1st.  Show  how  many  values  y  has  i?i  gerieral,  and 
whether  they  are  equal  or  unequal;  2d.  For  what  particular 
value  or  values  of  x,  y  has  but  one  value ;  3d.  For  what 
values  of  x,  y  is  real,  and  for  what  imaginary;  4th.  For 
what  values  o^  x,  y  is  +,  and  for  what  — ;  5th.  Also 
determine  what  values  of  x  render  y  infinite : 

5.  y^^2xy—2x^—4.y—x-^10  =  0;* 

6.  y^—2xy-{-2x^—2y  +  2x  =  0  ; 

7.  y^-]-2xy  +  x^—Gy-^d  =  0; 

8.  y^-{-2xy-\-Sx^—4:X  =  0; 

9.  y'^—2xy-{-3x^-i-2y—4:X—3  =  0; 

10.  y^-\-2xy—3x^—4:X  =  0; 

11.  y^—2xy-\-x^-{-x  =  0; 

*  In  all  cases  solve  the  equation  for  y  in  the  first  place.    In  this  example 
y  =  — x  +  2  ±  V3a;*— 3a;-^. 


430         IKTERPRETATIOK  OF  EQUATIONS. 

12.  y'^—2xy  +  x^—4.y  +  x  +  4.  =  {)', 

13.  y^—2xy  +  x^  +  2y  +  l  =  0; 

14.  2/2— 2ic2— 2^  +  6a;— 3  =  0  ; 

15.  y'^—%xy—dx^-2y  +  1x—l  =  0; 

16.  y^—2xy—2  =  0; 

17.  y^—2xy-]-2y-{-^x—S  =  0  ; 

18.  4?/2+4^2_|_2t/_3a;  +  12  =  0  ; 

19.  3^2_8a;2  =  12  ; 

20.  12?/2  +  4x2  =  20; 

21.  x^  +  y^  =  16; 

22.  x^—y^  =z  20. 

198,  Arithmetical  Interpretations  of  Nega- 
tive and  Imaginary  Solutions, 

1.  A  is  20  years  old,  and  B  16.  When  will  A  be  twice  as 
old  as  B  ? 

SuG's.— We  have  20  +  a?  =  2  (16  +  a*) ;  whence  a;  =  — 12.  The  arith- 
metical interpretation  of  this  result  is  that  A  will  never  be  twice  as  old 
as  B,  but  that  he  was  twice  as  old  12  years  ago,  i.  e,,  when  he  was  8 
and  B  4. 

2.  A  is  «  years  old,  and  B  b.  When  will  A  be  w  times  as 
old  as  B?  For  ^>1  what  are  the  possible  relative  values 
of  a  and  &  consistently  with  the  arithmetical  sense  of  the 
problem  ?  Interpret  for  a'>nb,  a  =z  nh,  a<^nb  when  w>l. 
Also  for  n=^l,  aynb,  a<Cnb,  and  a  =  nb. 

3.  Two  couriers,  A  and  B,  are  traveling  the  same  road  in 
the  same  direction,  the  former  at  rate  a,  the  latter  at  rate  h. 
They  are  at  two  places  c  miles  apart  at  the  same  time. 
Where  and  when  are  they  together  ? 

Solution  and  Discussion. — Let  XY  represent  the  road  which  the 
couriers  are  traveling  in  the  direction  from  X  to  Y,  and  A  and  B 
the  stations  which  they  pass  at  the  same  time,  A  being  at  A  when  B 
is  at  B,,  and  D  or  D'  the  place  at  which  they  are  together.     Call  the 


INTERPRETATION  OF  EQUATIONS.         431 

distance  from  B  to  the  place  at  whicli  they  are  together  ±x,  +x  when 
D  is  beyond  B,  and  —x  when  it  is  on  the  hither  side  of  A  and  B,  as  at 


A^ 


D'.  Then  the  distance  from  A  to  the  point  at  which  they  are  together 
is  c  +  (±x).  Now  disregarding  the  essential  sign  of  x,  and  leaving  it 
to  be  determined  in  the  sequel,  we  have 

Distance  A  travels  from  A  =  c  +  x, 
Distance  B  travels  from  B  —       x  ; 

Time  from  passing  A  and  B  to  the  time  they  are  together 

and  -.     But  these  are  equal.     Hence  we  are  to  discuss  the  equation. 

c  +  x      X  be  -  ac 

=  -,  or  a;  :=  - — ^  ,  and  c  +  x  = =■ . 

a        0  a—o  a—o 


The  points  to  be  noticed  in  the  discussion  are,  (1)  when  a>h,  (2) 
when  a<b,  (3)  when  a  =  b,  c  being  greater  than  0  in  each  case  but 
not  00 .    Also  the  like  cases  when  c  =  0. 

When  c<0  but  not  oo . 

We  have,  for  a>b,  x  positive,  which  shows  that  the  point  at  which 
they  are  together  is  at  the  right  of  B,  i.  e.,  in  the  direction  which  they 

are  traveling.     The  time,  tIoi" )>  is  positive,  which  shows  that 

they  are  together  after  passing  A  and  B, 

For  a<b,xis  negative,  and  c  +  x,  which  equals  — -r  ,  is  also  nega- 

C(i — 0 

tive.  This  shows  that  they  were  together  at  a  point  at  the  left  of  A, 
that  is,  before  they  reached  the  stations  A  and  B.  With  this  the 
expressions  for  the  time  also  agree.     Thus  -  becomes  — -,  and 


is  also  negative,  since  in  this  case  x>c. 


432  Il!TTERPKETATION    OF    EQUATIONS. 

inTi  T  ho       he  J        ^  ac        ac 

When    a  =  T),    x  =  — -  =  -  =  oo ,   and.    c  +  x  =  — -=■  ==  -  =  go 
a—h      0  a—h      0 

wliicli  indicates  tliat  they  are  never  together. 
When  c  =  0. 

In  this  case  x  = =  0,  and  c  +  x  =  — -  =  0,  for  a  and  6  unequal, 

a—b  a—o 

indicating  that  they  are  together  when  they  are  at  A  and  B.     This  is 

evidently  correct,  since  A  and  B  coincide  in  this  case.     When  a  =  b, 

X  =  — ^  =  K  >  aiid  c  +  x  =  -p:,  which  shows  that  they  are  always  to- 
a—o      0  0 

gether,  -  being  a  symbol  of  indetermination  which  in  this  instance 

may  have  any  value  whatever,  as  we  see  from  the  nature  of  the 
problem. 

194,  ScH. — The  student  should  not  understand  that  the  symbol 

-  always  indicates  that  the  quantity  which  takes  this  form  has  an 

indefinite  number  of  values.  It  is  frequently  so,  but  not  necessarily. 
The  indetermination  may  be  only  apparent,  and  what  the  value  of  the 
expression  is  must  be  determined  from  other  considerations.  The 
Calculus  affords  the  most  elegant  general  methods  of  evaluating  such 
expressions.     But  the  simple  processes  of  Algebra  will  often  suffice. 

Thus  for  x  =  l,  T^^  =^,    But  zr^  =  1+x  +  x^,  which,  for  x  =  l, 
1—x       0  1—x 

1—x^ 
is  3.    Hence  — '—  =  3,  for  x  =  1.     Here  the  apparent  indeterminar 
1—x 

tion  arises  from  the  fact  that  the  particular  assumption  (that  a;  =  1) 

causes  the  two  quantities  between  which  we  wish  the  ratio,  viz.,  the 

numerator  and  denominator,  to  disappear.     Let  the  student  find  that 

^-I^ =31  for  x  =  l.     (See  also   190,  3d  part  of  demon- 

1—x+x^ — x^ 

stration.) 

4.  Two  couriers  starting  at  the  same  time  from  the  two 
points  A  and  B,  c  miles  apart,  travel  foivard  eacli  other  at 
the  rates  a  and  b  respectively.  Discuss  the  problem  with 
reference  to  the  place  and  time  of  meeting.  (Consider  when 
ayb,  a<J)y  and  a  =ib.) 


IKTERPEETATIOK   OF   EQUATIOiq-S.  433 

6.  Two  couriers,  A  and  B,  are  traveling  the  same  road  in 
the  same  direction,  the  former  at  rate  a,  and  the  latter  n 
times  as  fast.  They  are  at  two  places  c  miles  apart  at  the 
same  time.  Discuss  the  problem  with  reference  to  place 
and  time  of  meeting  as  in  Ex.  3,  adding  the  considerations, 
n  >  1,  ^i  <  1,  ?z  ==  1,  w  =  0. 

6.  Divide  10  into  two  parts  whose  product  shall  be  40. 

Solution  and  Discussion. — Let  x  and  y  be  the  parts,  then  x  -^  y 
=  10,  xy  =  40,  and  x  =  5  ±  ^  — 15,  y  =  5  t  ^  —  15.  These  re- 
sults we  find  to  be  imaginary.  This  signifies  that  the  problem  in  its 
arithmetical  signification  is  impossible  :  this  indeed  is  evident  on  the 
face  of  it.  But,  although  impossible  in  the  arithmetical  sense,  the 
values  thus  found  do  satisfy  "Chq  formal,  or  algebraic  sense.  Thus  the 
8um  of  5  +  /y/—  15  and  5  —  /\/—  15  is  10,  and  the  product  40. 

7.  The  sum  of  two  numbers  is  required  to  be  a,  and  the 
product  h :  what  is  the  maximum  value  of  5  which  will  ren- 
der the  problem  possible  in  the  arithmetical  sense  ?  What 
are  the  parts  for  this  value  of  5  ? 

8.  Divide  a  into  two  parts,  such  that  the  sum  of  their 
squares  shall  be  a  minimum, 

SuG's.— Let  X  and  a  —  «  be  the  parts,  and  m  the  minimum  sum. 
Then 

a;2  +  (<x-  xf  =  2x^  -2ax  +  a^=  m; 


whence  x  —  ^a  ±  \^2m  —  a^.  From  this  we  see  that  if  2m>a?  x  is 
imaginary.  Hence  the  least  value  which  we  can  have  is  2m  =  a?,  or 
m  =  \a\ 

9.  Divide  a  into  two  parts,  such  that  the  sum  of  the 
square  roots  shall  be  a  maximum. 

10.  Let  d  be  the  diiference  between  two  numbers:  re- 
quired that  the  square  of  the  greater  divided  by  the  less 
shall  be  a  minimum. 


434  Il^TERPRETATION   OF   EQUATIONS. 

11.  Let  a  and  l  be  two  numbers  of  which  a  is  the  greater, 
to  find  a  number  such  that  if  a  be  added  to  this  number, 
and  h  be  subtracted  from  it,  the  product  of  this  sum  and 
this  difference,  divided  by  the  square  of  the  number,  shall 
be  a  maximum. 

Sug's. — Let  n  be  the  number,  and  m  the  required  maximum  quo- 

x-     i      mi        ,     XT  j-^-         n^  ■¥  {a  —  h)n  —  ah  , 

tient.     Then  by  the  conditions  ~ •  =  m,  whence  we 

find 


a  —  h  ^a?  +  2a&  +  &^  —  ^ahm 

2(1  -  m)  ^  3(1  -  m) 


From  this  we  see  that  the  greatest  value  which  m  can  have  and  ren- 

-j-^  .    This  gives  n  =  —  ^r—. x  = . 

iab  ^  2{1  —  m)       a  —  b 


,              ,  .             (a  +  bf      mi .      .                       a  —  b           2ab 
der  n  real  is  m  =  ^  .  /  .    This  gives  n  = — — ^  =  


12.  To  find  the  point  on  a  line  passing  through  two 
lights  at  which  the  illumination  will  be  the  same  from  each 
light. 

Solution. — Let  A  and  B  be  the  two  lights,  and  XY  the  line  passing 


-e- 


D  A  B  D 

through  them.  Let  a  be  the  intensity  of  the  light  A  at  a  unit's  dis- 
tance from  it,  6  the  intensity  of  B  at  a  unit's  distance  from  it,  c  the 
distance  between  the  two  lights,  as  AB,  and  x  the  distance  of  the  point  of 
equal  illumination  from  the  light  A,  as  AD  (or  AD').  Then,  as  we  learn 
from  Physics  that  the  illuminating  effect  of  a  light  varies  inversely  as 
the  square  of  the  distance  from  it,  we  have  for  the  illumination  of  the 

point  D  by  light  A  —  ,  and  for  the  illumination  of  the  same  point  by 


light  B,  ^^  -_     g .    But  by  the  conditions  of  the  problem  these  efiects 


(c  -  xf 
are  equal ;  hence  we  have  the  equation  to  be  discussed ;  viz., 

ft  _       b 


IKTERPRETATIOK   OF   EQUATIONS.  435 

or  .  1  =  — 2:-_  ;  or  -  =  -^ ^^ — » 

or,  finally,        a;  =  c  — ~- —  ,  and  x  —  - 


which  are  the  values  of  x  to  be  discussed. 
Discussion. — I.  Let  c  he  finite  and  >  0. 

1.  When  a  >'b,x  =  c  —~ =-  >  t  c,   since  — =.-^- 


a  >b.  This  is  as  it  should  be,  since  for  a  >  h  the  point  of  equal 
illumination  will  evidently  be  nearer  to  B  than  to  A.    Again,  the  other 

value  of  X  gives  x  =  c  — =  >  c,  since  — = —is  +  and  >  1, 

y^a  —  ^b  ysja  —  y& 

when  a  >b.     Hence  we  learn  that  there  is  a  point  beyond  B,  as  at  D', 

where  the  illumination  is  the  same  from  each  light. 

If  we  assume  ^/a  =  3y^&,  AD  =  f  c,  and  AD'  =  2c. 

2.  It  is  evidently  unnecessary  to  consider  the  case  when  a<b,  since 
this  would  only  situate  the  points  of  equal  illumination  with  refer- 
ence to  A  as  the  preceding  discussion  does  with  reference  to  B. 

3.  Whena  =  &, 

X  =  c  — ^ _=  ^, 

^Ja-\-^b 

\fab  \/a       . 

smce,  — r-^^ =  =  —5^  =  i« 

This  is  as  it  should  be,  since  it  is  evident  that  in  this  case  the  point 
of  equal  illumination  is  midway  between  the  lights.  Again,  for  the 
second  value  of  Xy  we  have 

\/a 

X  =  C J^ —  GO  . 

^a—\^b 

This  is  also  evidently  correct ;  for  when  the  lights  are  of  equal  inten- 
sity  there  can  be  no  point  beyond  B,  for  example,  at  wliich  the  illu- 


436  PERMtJTATIOl^S. 

mination  from  A  will  be  equal  to  that  from  B,  except  when  a;  =  00, 
for  which  the  illumination  is  0  for  each  light.  [Let  the  student  give 
the  reason.] 

II.  When  c  =  0.      In  this  case  the  original  equation  —  =  ; 

x^      {c—xf 

becomes  -^  =  -5 ,  whence  a  =  &.     We  then  have 

x  =  c  — ^—^  =  0 ; 

^Ja  +  ^b 

and  aj  =  c  —-J- — -  =  —^ — -  =  ^. 

The  former  shows  that  there  is  a  point  of  equal  illumination  where 
the  lights  are  (when  c  =  0  they  are  together),  and  the  latter  shows 
that  any  point  in  the  line  is  equally  illuminated  by  each  light.  Both 
these  conclusions  are  evidently  correct 


SECTION    VI. 

PE  R  M  U  TATI  0  N  S. 

195,  Combinations  are  the  different  groups  which 
can  be  made  of  m  things  taken  ?z  in  a  group,  n  being  less 
than  m, 

III. — Taking  the  5  letters  a,  h,  c,  d,  e,  we  have  the  10  following 
combinations  when  the  letters  are  taken  3  in  a  group,  or,  as  it  is 
usually  expressed,  taken  3  and  3 :  abc,  abd,  abe,  acd,  ace,  ade,  bed,  bee, 
bde,  cde.  Taken  2  and  3,  we  have  the  following  10  combinations  :  ab, 
ac,  ad,  ae,  be,  bd,  be,  cd,  ce,  de.  It  is  to  be  noticed  that  no  two  combina- 
tions contain  the  sa7ne  letters;  i.  e.,  they  are  different  groups. 

196,  JPermutations  are  the  different  orders  in  which 
things  can  succeed  each  other. 


PERMUTATIONS.  437 

III. — Thus  the  two  letters  a,  b,  have  the  two  permutations  a5,  bd. 
The  three  letters  a,  b,  c  have  the  6  permutations  abc,  acb,  cab,  bac, 
bca,  cba. 

107,  Ar7*auf/ements  are  permutations  of  com- 
binations. 

III. — Taking  the  10  combinations  of  5  letters  taken  3  and  3,  and 
permuting  each  combination,  we  get  the  arrangements  of  5  letters 
taken  3  and  3.  Thus  the  combination  abc  gives  the  6  arrangements 
abc,  acb,  cab,  bac,  bca,  cba.  In  like  manner  each  of  the  10  combina- 
tions  of  5  letters  taken  3  and  3  will  give  6  arrangements  ;  whence,  in 
all,  5  letters  taken  3  and  3  have  60  arrangements. 

198,  JProp* — Tlie  number  of  Arrangements  of  m  things 
taken  n  and  n  is 

m  (m— 1)  (m— 2)  (m— 3) (m— n  +  1). 

Dem. — Let  us  consider  the  number  of  arrangements  which  can  be 
made  of  the  m  letters  a,  b,  c,  d,  etc.,  taken  2  and  2.  Letting  a  stand 
first,  we  can  have  ab,  ac,  ad,  etc.,  to  m— 1  arrangements.  Letting  b 
stand  first,  we  can  have  ba,  be,  bd,  etc.,  to  m— 1  arrangements  in  each 
case,  or  m{m—l)  arrangements  in  all. 

Again,  each  of  these  m{ni—  1)2  and  2  arrangements  will  give 
m— 2  arrangements  3  and  3,  by  placing  before  it  each  of  the  letters 
not  involved  in  it.  Thus  we  have  m{in—l)  {m—2)  arrangements  of 
m  letters  taken  3  and  3. 

Once  more,  each  of  these  m  (m—l)  {m—2)  d  and  3  arrangements 
will  give  m—S  arrangements  4  and  4,  by  placing  before  it  each  of  the 
letters  riot  involved  in  it.  Thus  we  have  m  (m—l)  {m—2)  (m— 3) 
arrangements  of  m  letters  taken  4  and  4. 

Finally,  we  observe  the  law;  i.e.,  the  number  of  arrangements 
is  equal  to  the  continued  product  of  m  {m—l)  {m—2)  (wi— 3)  .  .  .  . 
{m—{n—l)\  0Tm{7n—l)  {m—2)  {m—d) ....  {m—n  +  1). 

199,  Cor.   1.  —  T/ie  number  of  Permutations  of  m 

things  is 

1.2.3.4 m. 

Tliis  is  evident  since  arrangements  become  permutations  when  the 
number  in  a  group  is  equal  to  the  whole  number  considered ;  i.  e.y 
when  n  =  m. 


438  PERMUTATIONS. 

200,  Cor.  2. — If  p  of  the  m  letters  are  alike  {as  each  a), 
q  others  alike,  r  others  alike,  etc,  the  mtmher  of  permuta- 
tions is 

1.2-3.4 m 

[p  X  [q  X  |r  X  etc. 

Thus  consider  the  permutations  of  a,  b,  c,  d,  viz.,  abed,  bacd,  acdb, 

bcda,  acbd,  bead,  abdc,  bade,  add),  bdca,  etc.     Suppose  b  to  become  a, 

then  since  for  any  particular  position  of  c  and  d,  as  in  abed,  there  are 

as  many  permutations  of  the  four  letters  as  there  can  be  permutations 

of  the  two  letters  a  and  b,  viz.,  1x2;    if  &  becomes  a  there  will  be 

1x2  fewer  permutations  when  these  two  letters  are  alike  than  when 

*i  A'^        .    •        1-3-3-4 

they  are  different,  i.  e.,  —      —  • 

1  •  4i 

So,  in  general,  if  p  of  the  letters  are  alike,  there  will  be  1  •  2  •  3  .  .  , 
p,  or  [£  fewer  permutations  than  if  they  are  all  different,  etc. 

201.  Cor.  3. — Tlie  mimher  of  Combinations  of  m  tilings 
taken  n  and  n  is 

m  (m  — 1)  (m  — 2)  (m— 3) (m— n  +  1) 

1^2.3-4 n 

Since  arrangements  are  permutations  of  combinations,  the  number 
of  arrangements  of  m  things  taken  n  and  /«.  is  equal  to  the  number  of 
combinations  of  m  things  taken  n  and  n  multiplied  by  the  number  of 
permutations  of  n  things.  Hence  the  number  of  combinations  is  equal 
to  the  number  of  arrangements  of  m  things  taken  n  and  n  divided  by 
the  number  of  permutations  of  n  things. 


EXAM  PLE  S. 

1.  How  many  permutations  can  be  made  of  the  letters  in 
the  word  marble?  Of  those  in  home?  Of  those  in 
logarithms? 

2.  How  many  arrangements  can  be  made  of  10  colors 
taken  3  and  3  ?  Of  7  colors  taken  2  and  2  ?  Taken  3 
and  3  ?  4  and  4  ?  5  and  5  ?  G  and  6  ?  7  and  7  ?  How 
many  mixtures  in  each  case,  irrespective  of  proportions  ? 


permutatio:n'S.  439 

3.  How  many  different  products  can  be  made  from  the 
9  digits  taken  2  and  2  ?  3  and  3  ?  4  and  4  ?  5  and  5  ? 
6  and  6  ?     7  and  7  ?     8  and  8  ?     9  and  9  ? 

4.  How  many  different  numbers  can  be  represented  by 
the  9  digits  taken  2  and  2  ?     3  and  3  ?     4  and  4  ?  etc. 

5.  In  a  certain  district  3  representatives  are  to  be  elected, 
and  there  are  6  candidates.  In  how  many  different  ways 
may  a  ticket  be  made  up.? 

6.  There  are  7  chemical  elements  which  will  unite  with 
each  other.  How  many  ternary  compounds  can  be  made 
from  them  ?     How  many  binary  ? 

7.  How  many  different  sums  of  money  can  be  paid  with 

1  cent,  1  3-cent  piece,  1  5-cent  piece,  1  dime,  1  15-cent 
piece,  1  25-cent  piece,  and  1  50-cent  piece  ? 

SuG.— If  taken  1  and  1,  how  many  ?  If  2  and  2,  how  many  ?  If  3 
and  3,  etc.  ?    How  many  in  all  ? 

8.  In  how  many  ways  can  12  ladies  and  12  gentlemen 
arrange  themselves  in  couples  ? 

9.  If  you  are  to  select  7  articles  out  of  12,  how  many 
different  choices  have  you  ? 

10.  How  many  different  sums  can  be  made  from  1,  2,  3, 
4,  5,  6,  taken  2  and  2  ? 

11.  How  many  permutations  can  be  made  from  the 
letters  in  the  word  possessions?  (See  200,)  How 
many  from  the  letters  in  the  word  consisten- 
cies? 

12.  How  many  different  signals  can  be  made  with  10 
different-colored  flags,  by  displaying   them   1   at  a   time, 

2  at  a  time,  3  at  a  time,  etc.,  the  relative  position  of  the 
flags  with  reference  to  each  other  not  being  taken  into 
account  ? 


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The  whole  subject  in  Tivo  Books, 

These  hooks  are  the  most  simple,  the  most  practical,  and  ifest 
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For  those  desiring  to  pursue  the  study  of  Physical  Geography, 
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Sheldon  d-  Co???pa??J^'s  Text-liooks 


HISTOEIES  OF  THE  UNITED  STATES. 

By  Benson  J.  Lossing,  author  of  "  Field-Book  of  the  Revolu- 
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Lossing^ s  School  History,    383  pages. 

Containing  the   National   Constitution,  Declaration  of 
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viduals and  governments.  The  Third  delineates  the  progress  of  all  the  Setile- 
«<f^HY«  until  colonial  governments  were  formed.  The  Fourth  tells  the  story 
of  these  Colonies  from  their  infancy  to  maturity,  and  illustrates  the  continual 
development  of  democratic  ideas  and  repiiblican  tendencies  which  finally 
resulted  in  a  political  confederation.  The  Fifth  has  a  full  account  of  the  im- 
portant events  of  the  War  for  Independence  :  and  the  Sixth  gives  a  con- 
cise History  of  the  Itepublic  from  its  formation  to  the  present  time. 


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The  entire  series  is  characterized  by  chasteness  and  clearness  of  style,  accuracy 
of  statement,  beauty  of  typography,  and  fullness  of  illustration.  The  author 
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history,  and  his  ability  and  reputation  are  a  sufficient  guarantee  that  the  work 
has  been  thoroughly  done,  and  a  series  of  histories  produced  that  will  be  In- 
valuable in  training  and  educating  the  youth  of  our  country. 


V.V 


